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Partial Fractions

Integrate a rational function using the method of partial fractions.

Nonrepeated Linear Factors

If \(Q(x)\) can be factored as \(({a}_{1}x+{b}_{1})({a}_{2}x+{b}_{2})\text{\ldots }({a}_{n}x+{b}_{n}),\) where each linear factor is distinct, then it is possible to find constants \({A}_{1},{A}_{2}\text{,\ldots }\ {A}_{n}\) satisfying

\[\frac{P(x)}{Q(x)}=\frac{{A}_{1}}{{a}_{1}x+{b}_{1}}+\frac{{A}_{2}}{{a}_{2}x+{b}_{2}}+\cdots +\frac{{A}_{n}}{{a}_{n}x+{b}_{n}}.\]

The proof that such constants exist is beyond the scope of this course.

In this next example, we see how to use partial fractions to integrate a rational function of this type.

In the next example, we integrate a rational function in which the degree of the numerator is not less than the degree of the denominator.

Example

Try it.

Evaluate \(\int \frac{{x}^{2}+3x+1}{{x}^{2}-4}dx.\)

Solution

Since \(\text{degree}({x}^{2}+3x+1)\ge \text{degree}({x}^{2}-4),\) we must perform long division of polynomials. This results in

\[\frac{{x}^{2}+3x+1}{{x}^{2}-4}=1+\frac{3x+5}{{x}^{2}-4}.\]

Next, we perform partial fraction decomposition on \(\frac{3x+5}{{x}^{2}-4}=\frac{3x+5}{(x+2)(x-2)}.\) We have

\[\frac{3x+5}{(x-2)(x+2)}=\frac{A}{x-2}+\frac{B}{x+2}.\]

Thus,

\[3x+5=A(x+2)+B(x-2).\]

Solving for \(A\) and \(B\) using either method, we obtain \(A=11\text{/}4\) and \(B=1\text{/}4.\)

Rewriting the original integral, we have

\[\int \frac{{x}^{2}+3x+1}{{x}^{2}-4}dx=\int (1+\frac{11}{4}\cdot \frac{1}{x-2}+\frac{1}{4}\cdot \frac{1}{x+2})dx.\]

Evaluating the integral produces

\[\int \frac{{x}^{2}+3x+1}{{x}^{2}-4}dx=x+\frac{11}{4}\text{ln}|x-2|+\frac{1}{4}\text{ln}|x+2|+C.\]

As we see in the next example, it may be possible to apply the technique of partial fraction decomposition to a nonrational function. The trick is to convert the nonrational function to a rational function through a substitution.

Example

Try it.

Evaluate \(\int \frac{\text{cos}\ x}{{\text{sin}}^{2}x-\text{sin}\ x}dx.\)

Solution

Let’s begin by letting \(u=\text{sin}\ x.\) Consequently, \(du=\text{cos}\ x\ dx.\) After making these substitutions, we have

\[\int \frac{\text{cos}\ x}{{\text{sin}}^{2}x-\text{sin}\ x}dx=\int \frac{du}{{u}^{2}-u}=\int \frac{du}{u(u-1)}.\]

Applying partial fraction decomposition to \(1\text{/}u(u-1)\) gives \(\frac{1}{u(u-1)}=-\frac{1}{u}+\frac{1}{u-1}.\)

Thus,

\[\begin{array}{ll}\int \frac{\text{cos}\ x}{{\text{sin}}^{2}x-\text{sin}\ x}dx & =\text{-}\text{ln}|u|+\text{ln}|u-1|+C \\ & =\text{-}\text{ln}|\text{sin}\ x|+\text{ln}|\text{sin}\ x-1|+C.\end{array}\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Repeated Linear Factors

For some applications, we need to integrate rational expressions that have denominators with repeated linear factors—that is, rational functions with at least one factor of the form \({(ax+b)}^{n},\) where \(n\) is a positive integer greater than or equal to \(2.\) If the denominator contains the repeated linear factor \({(ax+b)}^{n},\) then the decomposition must contain

\[\frac{{A}_{1}}{ax+b}+\frac{{A}_{2}}{{(ax+b)}^{2}}+\cdots +\frac{{A}_{n}}{{(ax+b)}^{n}}.\]

As we see in our next example, the basic technique used for solving for the coefficients is the same, but it requires more algebra to determine the numerators of the partial fractions.

Example

Try it.

Evaluate \(\int \frac{x-2}{{(2x-1)}^{2}(x-1)}dx.\)

Solution

We have \(\text{degree}(x-2)<\text{degree}({(2x-1)}^{2}(x-1)),\) so we can proceed with the decomposition. Since \({(2x-1)}^{2}\) is a repeated linear factor, include \(\frac{A}{2x-1}+\frac{B}{{(2x-1)}^{2}}\) in the decomposition. Thus,

\[\frac{x-2}{{(2x-1)}^{2}(x-1)}=\frac{A}{2x-1}+\frac{B}{{(2x-1)}^{2}}+\frac{C}{x-1}.\]

After getting a common denominator and equating the numerators, we have

\[x-2=A(2x-1)(x-1)+B(x-1)+C{(2x-1)}^{2}.\]

We then use the method of equating coefficients to find the values of \(A,\) \(B,\) and \(C.\)

\[x-2=(2A+4C){x}^{2}+(-3A+B-4C)x+(A-B+C).\]

Equating coefficients yields \(2A+4C=0,\) \(-3A+B-4C=1,\) and \(A-B+C=-2.\) Solving this system yields \(A=2,\) \(B=3,\) and \(C=-1.\)

Alternatively, we can use the method of strategic substitution. In this case, substituting \(x=1\) and \(x=1\text{/}2\) into easily produces the values \(B=3\) and \(C=-1.\) At this point, it may seem that we have run out of good choices for \(x,\) however, since we already have values for \(B\) and \(C,\) we can substitute in these values and choose any value for \(x\) not previously used. The value \(x=0\) is a good option. In this case, we obtain the equation \(-2=A(-1)(-1)+3(-1)+(-1){(-1)}^{2}\) or, equivalently, \(A=2.\)

Now that we have the values for \(A,\) \(B,\) and \(C,\) we rewrite the original integral and evaluate it:

\[\begin{array}{ll}\int \frac{x-2}{{(2x-1)}^{2}(x-1)}dx & =\int (\frac{2}{2x-1}+\frac{3}{{(2x-1)}^{2}}-\frac{1}{x-1})dx \\ & =\text{ln}|2x-1|-\frac{3}{2(2x-1)}-\text{ln}|x-1|+C.\end{array}\]

The General Method

Now that we are beginning to get the idea of how the technique of partial fraction decomposition works, let’s outline the basic method in the following problem-solving strategy.

Partial Fraction Decomposition

To decompose the rational function \(P(x)\text{/}Q(x),\) use the following steps:

Make sure that \(\text{degree}(P(x))<\text{degree}(Q(x)).\) If not, perform long division of polynomials.
Factor \(Q(x)\) into the product of linear and irreducible quadratic factors. An irreducible quadratic is a quadratic that has no real zeros.
Assuming that \(\text{deg}(P(x))<\text{deg}(Q(x)),\) the factors of \(Q(x)\) determine the form of the decomposition of \(P(x)\text{/}Q(x).\)
1. If \(Q(x)\) can be factored as \(({a}_{1}x+{b}_{1})({a}_{2}x+{b}_{2})\text{\ldots }({a}_{n}x+{b}_{n}),\) where each linear factor is distinct, then it is possible to find constants \({A}_{1},{A}_{2},...{A}_{n}\) satisfying
  \[\frac{P(x)}{Q(x)}=\frac{{A}_{1}}{{a}_{1}x+{b}_{1}}+\frac{{A}_{2}}{{a}_{2}x+{b}_{2}}+\cdots +\frac{{A}_{n}}{{a}_{n}x+{b}_{n}}.\]
2. If \(Q(x)\) contains the repeated linear factor \({(ax+b)}^{n},\) then the decomposition must contain
  \[\frac{{A}_{1}}{ax+b}+\frac{{A}_{2}}{{(ax+b)}^{2}}+\cdots +\frac{{A}_{n}}{{(ax+b)}^{n}}.\]
3. For each irreducible quadratic factor \(a{x}^{2}+bx+c\) that \(Q(x)\) contains, the decomposition must include
  \[\frac{Ax+B}{a{x}^{2}+bx+c}.\]
4. For each repeated irreducible quadratic factor \({(a{x}^{2}+bx+c)}^{n},\) the decomposition must include
  \[\frac{{A}_{1}x+{B}_{1}}{a{x}^{2}+bx+c}+\frac{{A}_{2}x+{B}_{2}}{{(a{x}^{2}+bx+c)}^{2}}+\cdots +\frac{{A}_{n}x+{B}_{n}}{{(a{x}^{2}+bx+c)}^{n}}.\]
5. After the appropriate decomposition is determined, solve for the constants.
6. Last, rewrite the integral in its decomposed form and evaluate it using previously developed techniques or integration formulas.

Simple Quadratic Factors

Now let’s look at integrating a rational expression in which the denominator contains an irreducible quadratic factor. Recall that the quadratic \(a{x}^{2}+bx+c\) is irreducible if \(a{x}^{2}+bx+c=0\) has no real zeros—that is, if \({b}^{2}-4ac<0.\)

Example

Try it.

Evaluate \(\int \frac{2x-3}{{x}^{3}+x}dx.\)

Solution

Since \(\text{deg}(2x-3)<\text{deg}({x}^{3}+x),\) factor the denominator and proceed with partial fraction decomposition. Since \({x}^{3}+x=x({x}^{2}+1)\) contains the irreducible quadratic factor \({x}^{2}+1,\) include \(\frac{Ax+B}{{x}^{2}+1}\) as part of the decomposition, along with \(\frac{C}{x}\) for the linear term \(x.\) Thus, the decomposition has the form

\[\frac{2x-3}{x({x}^{2}+1)}=\frac{Ax+B}{{x}^{2}+1}+\frac{C}{x}.\]

After getting a common denominator and equating the numerators, we obtain the equation

\[2x-3=(Ax+B)x+C({x}^{2}+1).\]

Solving for \(A,B,\) and \(C,\) we get \(A=3,\) \(B=2,\) and \(C=-3.\)

Thus,

\[\frac{2x-3}{{x}^{3}+x}=\frac{3x+2}{{x}^{2}+1}-\frac{3}{x}.\]

Substituting back into the integral, we obtain

\[\begin{array}{lllll}\int \frac{2x-3}{{x}^{3}+x}dx & =\int (\frac{3x+2}{{x}^{2}+1}-\frac{3}{x})dx & & & \\ & =3\int \frac{x}{{x}^{2}+1}dx+2\int \frac{1}{{x}^{2}+1}dx-3\int \frac{1}{x}dx & & & \text{Split up the integral.} \\ & =\frac{3}{2}\text{ln}|{x}^{2}+1|+2\ {\text{tan}}^{-1}x-3\ \text{ln}|x|+C. & & & \text{Evaluate each integral.}\end{array}\]

Note: We may rewrite \(\text{ln}|{x}^{2}+1|=\text{ln}({x}^{2}+1),\) if we wish to do so, since \({x}^{2}+1>0.\)

Example

Try it.

Find the volume of the solid of revolution obtained by revolving the region enclosed by the graph of \(f(x)=\frac{{x}^{2}}{{({x}^{2}+1)}^{2}}\) and the x-axis over the interval \([0,1]\) about the y-axis.

Solution

Let’s begin by sketching the region to be revolved (see ). From the sketch, we see that the shell method is a good choice for solving this problem.

The volume is given by

\[V=2\pi {\int }_{0}^{1}x\cdot \frac{{x}^{2}}{{({x}^{2}+1)}^{2}}dx=2\pi {\int }_{0}^{1}\frac{{x}^{3}}{{({x}^{2}+1)}^{2}}dx.\]

Since \(\text{deg}({({x}^{2}+1)}^{2})=4>3=\text{deg}({x}^{3}),\) we can proceed with partial fraction decomposition. Note that \({({x}^{2}+1)}^{2}\) is a repeated irreducible quadratic. Using the decomposition described in the problem-solving strategy, we get

\[\frac{{x}^{3}}{{({x}^{2}+1)}^{2}}=\frac{Ax+B}{{x}^{2}+1}+\frac{Cx+D}{{({x}^{2}+1)}^{2}}.\]

Finding a common denominator and equating the numerators gives

\[{x}^{3}=(Ax+B)({x}^{2}+1)+Cx+D.\]

Solving, we obtain \(A=1,\) \(B=0,\) \(C=-1,\) and \(D=0.\) Substituting back into the integral, we have

\[\begin{array}{ll}V & =2\pi \int _{0}^{1}\frac{{x}^{3}}{{({x}^{2}+1)}^{2}}dx \\ & =2\pi \int _{0}^{1}(\frac{x}{{x}^{2}+1}-\frac{x}{{({x}^{2}+1)}^{2}})dx \\ & =2\pi (\frac{1}{2}\text{ln}({x}^{2}+1)+\frac{1}{2}\cdot \frac{1}{{x}^{2}+1})|{}_{\begin{array}{l} \\ 0\end{array}}^{\begin{array}{l}1 \\ \end{array}} \\ & =\pi (\text{ln}\ 2-\frac{1}{2}).\end{array}\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Key Concepts

Partial fraction decomposition is a technique used to break down a rational function into a sum of simple rational functions that can be integrated using previously learned techniques.
When applying partial fraction decomposition, we must make sure that the degree of the numerator is less than the degree of the denominator. If not, we need to perform long division before attempting partial fraction decomposition.
The form the decomposition takes depends on the type of factors in the denominator. The types of factors include nonrepeated linear factors, repeated linear factors, nonrepeated irreducible quadratic factors, and repeated irreducible quadratic factors.

Partial Fractions

Express the rational function as a sum or difference of two simpler rational expressions.

Use the method of partial fractions to evaluate each of the following integrals.

Evaluate the following integrals, which have irreducible quadratic factors.

Use the method of partial fractions to evaluate the following integrals.

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals.

Use the given substitution to convert the integral to an integral of a rational function, then evaluate.

Solve the initial-value problem for x as a function of t.

Condensed — the full section is in OpenStax Calculus Volume 2.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Evaluate \(\int \frac{{x}^{2}+3x+5}{x+1}dx.\)

Откриј одговор.

Since \(\text{deg}({x}^{2}+3x+5)\ge \text{deg}(x+1),\) we perform long division to obtain
\[\frac{{x}^{2}+3x+5}{x+1}=x+2+\frac{3}{x+1}.\]
Thus,
\[\begin{array}{ll}\int \frac{{x}^{2}+3x+5}{x+1}dx & =\int (x+2+\frac{3}{x+1})dx \\ & =\frac{1}{2}{x}^{2}+2x+3\ \text{ln}|x+1|+C.\end{array}\]
Evaluate \(\int \frac{x-3}{x+2}dx.\)

Откриј одговор.

\(x-5\ \text{ln}|x+2|+C\)
Evaluate \(\int \frac{3x+2}{{x}^{3}-{x}^{2}-2x}dx.\)

Откриј одговор.

Since \(\text{deg}(3x+2)<\text{deg}({x}^{3}-{x}^{2}-2x),\) we begin by factoring the denominator of \(\frac{3x+2}{{x}^{3}-{x}^{2}-2x}.\) We can see that \({x}^{3}-{x}^{2}-2x=x(x-2)(x+1).\) Thus, there are constants \(A,\) \(B,\) and \(C\) satisfying
\[\frac{3x+2}{x(x-2)(x+1)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+1}.\]
We must now find these constants. To do so, we begin by getting a common denominator on the right. Thus,
\[\frac{3x+2}{x(x-2)(x+1)}=\frac{A(x-2)(x+1)+Bx(x+1)+Cx(x-2)}{x(x-2)(x+1)}.\]
Now, we set the numerators equal to each other, obtaining
\[3x+2=A(x-2)(x+1)+Bx(x+1)+Cx(x-2).\]
There are two different strategies for finding the coefficients \(A,\) \(B,\) and \(C.\) We refer to these as the method of equating coefficients and the method of strategic substitution.
Rule: Method of Equating Coefficients
Rewrite in the form
\[3x+2=(A+B+C){x}^{2}+(\text{-}A+B-2C)x+(-2A).\]
Equating coefficients produces the system of equations
\[\begin{array}{lll}A+B+C & = & 0 \\ -A+B-2C & = & 3 \\ -2A & = & 2.\end{array}\]
To solve this system, we first observe that \(-2A=2⇒A=-1.\) Substituting this value into the first two equations gives us the system
\[\begin{array}{lll}B+C & = & 1 \\ B-2C & = & 2.\end{array}\]
Multiplying the second equation by \(-1\) and adding the resulting equation to the first produces
\[-3C=1,\]
which in turn implies that \(C=-\frac{1}{3}.\) Substituting this value into the equation \(B+C=1\) yields \(B=\frac{4}{3}.\) Thus, solving these equations yields \(A=-1,\) \(B=\frac{4}{3},\) and \(C=-\frac{1}{3}.\)
It is important to note that the system produced by this method is consistent if and only if we have set up the decomposition correctly. If the system is inconsistent, there is an error in our decomposition.
Rule: Method of Strategic Substitution
The method of strategic substitution is based on the assumption that we have set up the decomposition correctly. If the decomposition is set up correctly, then there must be values of \(A,\) \(B,\) and \(C\) that satisfy for all values of \(x.\) That is, this equation must be true for any value of \(x\) we care to substitute into it. Therefore, by choosing values of \(x\) carefully and substituting them into the equation, we may find \(A,\) \(B,\) and \(C\) easily. For example, if we substitute \(x=0,\) the equation reduces to \(2=A(-2)(1).\) Solving for \(A\) yields \(A=-1.\) Next, by substituting \(x=2,\) the equation reduces to \(8=B(2)(3),\) or equivalently \(B=4\text{/}3.\) Last, we substitute \(x=-1\) into the equation and obtain \(-1=C(-1)(-3).\) Solving, we have \(C=-\frac{1}{3}.\)
It is important to keep in mind that if we attempt to use this method with a decomposition that has not been set up correctly, we are still able to find values for the constants, but these constants are meaningless. If we do opt to use the method of strategic substitution, then it is a good idea to check the result by recombining the terms algebraically.
Now that we have the values of \(A,\) \(B,\) and \(C,\) we rewrite the original integral:
\[\int \frac{3x+2}{{x}^{3}-{x}^{2}-2x}dx=\int (\text{-}\ \frac{1}{x}+\frac{4}{3}\cdot \frac{1}{(x-2)}-\frac{1}{3}\cdot \frac{1}{(x+1)})dx.\]
Evaluating the integral gives us
\[\int \frac{3x+2}{{x}^{3}-{x}^{2}-2x}dx=\text{-}\text{ln}|x|+\frac{4}{3}\text{ln}|x-2|-\frac{1}{3}\text{ln}|x+1|+C.\]
Evaluate \(\int \frac{{x}^{2}+3x+1}{{x}^{2}-4}dx.\)

Откриј одговор.

Since \(\text{degree}({x}^{2}+3x+1)\ge \text{degree}({x}^{2}-4),\) we must perform long division of polynomials. This results in
\[\frac{{x}^{2}+3x+1}{{x}^{2}-4}=1+\frac{3x+5}{{x}^{2}-4}.\]
Next, we perform partial fraction decomposition on \(\frac{3x+5}{{x}^{2}-4}=\frac{3x+5}{(x+2)(x-2)}.\) We have
\[\frac{3x+5}{(x-2)(x+2)}=\frac{A}{x-2}+\frac{B}{x+2}.\]
Thus,
\[3x+5=A(x+2)+B(x-2).\]
Solving for \(A\) and \(B\) using either method, we obtain \(A=11\text{/}4\) and \(B=1\text{/}4.\)
Rewriting the original integral, we have
\[\int \frac{{x}^{2}+3x+1}{{x}^{2}-4}dx=\int (1+\frac{11}{4}\cdot \frac{1}{x-2}+\frac{1}{4}\cdot \frac{1}{x+2})dx.\]
Evaluating the integral produces
\[\int \frac{{x}^{2}+3x+1}{{x}^{2}-4}dx=x+\frac{11}{4}\text{ln}|x-2|+\frac{1}{4}\text{ln}|x+2|+C.\]
Evaluate \(\int \frac{\text{cos}\ x}{{\text{sin}}^{2}x-\text{sin}\ x}dx.\)

Откриј одговор.

Let’s begin by letting \(u=\text{sin}\ x.\) Consequently, \(du=\text{cos}\ x\ dx.\) After making these substitutions, we have
\[\int \frac{\text{cos}\ x}{{\text{sin}}^{2}x-\text{sin}\ x}dx=\int \frac{du}{{u}^{2}-u}=\int \frac{du}{u(u-1)}.\]
Applying partial fraction decomposition to \(1\text{/}u(u-1)\) gives \(\frac{1}{u(u-1)}=-\frac{1}{u}+\frac{1}{u-1}.\)
Thus,
\[\begin{array}{ll}\int \frac{\text{cos}\ x}{{\text{sin}}^{2}x-\text{sin}\ x}dx & =\text{-}\text{ln}|u|+\text{ln}|u-1|+C \\ & =\text{-}\text{ln}|\text{sin}\ x|+\text{ln}|\text{sin}\ x-1|+C.\end{array}\]
Evaluate \(\int \frac{x+1}{(x+3)(x-2)}dx.\)

Откриј одговор.

\(\frac{2}{5}\text{ln}|x+3|+\frac{3}{5}\text{ln}|x-2|+C\)
Evaluate \(\int \frac{x-2}{{(2x-1)}^{2}(x-1)}dx.\)

Откриј одговор.

We have \(\text{degree}(x-2)<\text{degree}({(2x-1)}^{2}(x-1)),\) so we can proceed with the decomposition. Since \({(2x-1)}^{2}\) is a repeated linear factor, include \(\frac{A}{2x-1}+\frac{B}{{(2x-1)}^{2}}\) in the decomposition. Thus,
\[\frac{x-2}{{(2x-1)}^{2}(x-1)}=\frac{A}{2x-1}+\frac{B}{{(2x-1)}^{2}}+\frac{C}{x-1}.\]
After getting a common denominator and equating the numerators, we have
\[x-2=A(2x-1)(x-1)+B(x-1)+C{(2x-1)}^{2}.\]
We then use the method of equating coefficients to find the values of \(A,\) \(B,\) and \(C.\)
\[x-2=(2A+4C){x}^{2}+(-3A+B-4C)x+(A-B+C).\]
Equating coefficients yields \(2A+4C=0,\) \(-3A+B-4C=1,\) and \(A-B+C=-2.\) Solving this system yields \(A=2,\) \(B=3,\) and \(C=-1.\)
Alternatively, we can use the method of strategic substitution. In this case, substituting \(x=1\) and \(x=1\text{/}2\) into easily produces the values \(B=3\) and \(C=-1.\) At this point, it may seem that we have run out of good choices for \(x,\) however, since we already have values for \(B\) and \(C,\) we can substitute in these values and choose any value for \(x\) not previously used. The value \(x=0\) is a good option. In this case, we obtain the equation \(-2=A(-1)(-1)+3(-1)+(-1){(-1)}^{2}\) or, equivalently, \(A=2.\)
Now that we have the values for \(A,\) \(B,\) and \(C,\) we rewrite the original integral and evaluate it:
\[\begin{array}{ll}\int \frac{x-2}{{(2x-1)}^{2}(x-1)}dx & =\int (\frac{2}{2x-1}+\frac{3}{{(2x-1)}^{2}}-\frac{1}{x-1})dx \\ & =\text{ln}|2x-1|-\frac{3}{2(2x-1)}-\text{ln}|x-1|+C.\end{array}\]
Set up the partial fraction decomposition for \(\int \frac{x+2}{{(x+3)}^{3}{(x-4)}^{2}}dx.\) (Do not solve for the coefficients or complete the integration.)

Откриј одговор.

\(\frac{x+2}{{(x+3)}^{3}{(x-4)}^{2}}=\frac{A}{x+3}+\frac{B}{{(x+3)}^{2}}+\frac{C}{{(x+3)}^{3}}+\frac{D}{(x-4)}+\frac{E}{{(x-4)}^{2}}\)
Evaluate \(\int \frac{2x-3}{{x}^{3}+x}dx.\)

Откриј одговор.

Since \(\text{deg}(2x-3)<\text{deg}({x}^{3}+x),\) factor the denominator and proceed with partial fraction decomposition. Since \({x}^{3}+x=x({x}^{2}+1)\) contains the irreducible quadratic factor \({x}^{2}+1,\) include \(\frac{Ax+B}{{x}^{2}+1}\) as part of the decomposition, along with \(\frac{C}{x}\) for the linear term \(x.\) Thus, the decomposition has the form
\[\frac{2x-3}{x({x}^{2}+1)}=\frac{Ax+B}{{x}^{2}+1}+\frac{C}{x}.\]
After getting a common denominator and equating the numerators, we obtain the equation
\[2x-3=(Ax+B)x+C({x}^{2}+1).\]
Solving for \(A,B,\) and \(C,\) we get \(A=3,\) \(B=2,\) and \(C=-3.\)
Thus,
\[\frac{2x-3}{{x}^{3}+x}=\frac{3x+2}{{x}^{2}+1}-\frac{3}{x}.\]
Substituting back into the integral, we obtain
\[\begin{array}{lllll}\int \frac{2x-3}{{x}^{3}+x}dx & =\int (\frac{3x+2}{{x}^{2}+1}-\frac{3}{x})dx & & & \\ & =3\int \frac{x}{{x}^{2}+1}dx+2\int \frac{1}{{x}^{2}+1}dx-3\int \frac{1}{x}dx & & & \text{Split up the integral.} \\ & =\frac{3}{2}\text{ln}|{x}^{2}+1|+2\ {\text{tan}}^{-1}x-3\ \text{ln}|x|+C. & & & \text{Evaluate each integral.}\end{array}\]
Note: We may rewrite \(\text{ln}|{x}^{2}+1|=\text{ln}({x}^{2}+1),\) if we wish to do so, since \({x}^{2}+1>0.\)
Evaluate \(\int \frac{dx}{{x}^{3}-8}.\)

Откриј одговор.

We can start by factoring \({x}^{3}-8=(x-2)({x}^{2}+2x+4).\) We see that the quadratic factor \({x}^{2}+2x+4\) is irreducible since \({2}^{2}-4(1)(4)=-12<0.\) Using the decomposition described in the problem-solving strategy, we get
\[\frac{1}{(x-2)({x}^{2}+2x+4)}=\frac{A}{x-2}+\frac{Bx+C}{{x}^{2}+2x+4}.\]
After obtaining a common denominator and equating the numerators, this becomes
\[1=A({x}^{2}+2x+4)+(Bx+C)(x-2).\]
Applying either method, we get \(A=\frac{1}{12},B=-\frac{1}{12},\ \text{and}\ C=-\frac{1}{3}.\)
Rewriting \(\int \frac{dx}{{x}^{3}-8},\) we have
\[\int \frac{dx}{{x}^{3}-8}=\frac{1}{12}\int \frac{1}{x-2}dx-\frac{1}{12}\int \frac{x+4}{{x}^{2}+2x+4}dx.\]
We can see that
\(\int \frac{1}{x-2}dx=\text{ln}|x-2|+C,\) but \(\int \frac{x+4}{{x}^{2}+2x+4}dx\) requires a bit more effort. Let’s begin by completing the square on \({x}^{2}+2x+4\) to obtain
\[{x}^{2}+2x+4={(x+1)}^{2}+3.\]
By letting \(u=x+1\) and consequently \(du=dx,\) we see that
\[\begin{array}{lllll}\int \frac{x+4}{{x}^{2}+2x+4}dx & =\int \frac{x+4}{{(x+1)}^{2}+3}dx & & & \begin{array}{l}\text{Complete the square on the} \\ \text{denominator.}\end{array} \\ & =\int \frac{u+3}{{u}^{2}+3}du & & & \begin{array}{l}\text{Substitute}\ u=x+1,x=u-1, \\ \text{and}\ du=dx.\end{array} \\ & =\int \frac{u}{{u}^{2}+3}du+\int \frac{3}{{u}^{2}+3}du & & & \text{Split the numerator apart.} \\ & =\frac{1}{2}\text{ln}|{u}^{2}+3|+\frac{3}{\sqrt{3}}{\text{tan}}^{-1}\frac{u}{\sqrt{3}}+C & & & \text{Evaluate each integral.} \\ & =\frac{1}{2}\text{ln}|{x}^{2}+2x+4|+\sqrt{3}{\text{tan}}^{-1}(\frac{x+1}{\sqrt{3}})+C. & & & \begin{array}{l}\text{Rewrite in terms of}\ x\ \text{and} \\ \text{simplify.}\end{array}\end{array}\]
Substituting back into the original integral and simplifying gives
\[{\int }^{\text{}}\frac{dx}{{x}^{3}-8}=\frac{1}{12}\text{ln}|x-2|-\frac{1}{24}\text{ln}|{x}^{2}+2x+4|-\frac{\sqrt{3}}{12}{\text{tan}}^{-1}(\frac{x+1}{\sqrt{3}})+C.\]
Here again, we can drop the absolute value if we wish to do so, since \({x}^{2}+2x+4>0\) for all \(x.\)
Find the volume of the solid of revolution obtained by revolving the region enclosed by the graph of \(f(x)=\frac{{x}^{2}}{{({x}^{2}+1)}^{2}}\) and the x-axis over the interval \([0,1]\) about the y-axis.

Откриј одговор.

Let’s begin by sketching the region to be revolved (see ). From the sketch, we see that the shell method is a good choice for solving this problem.
The volume is given by
\[V=2\pi {\int }_{0}^{1}x\cdot \frac{{x}^{2}}{{({x}^{2}+1)}^{2}}dx=2\pi {\int }_{0}^{1}\frac{{x}^{3}}{{({x}^{2}+1)}^{2}}dx.\]
Since \(\text{deg}({({x}^{2}+1)}^{2})=4>3=\text{deg}({x}^{3}),\) we can proceed with partial fraction decomposition. Note that \({({x}^{2}+1)}^{2}\) is a repeated irreducible quadratic. Using the decomposition described in the problem-solving strategy, we get
\[\frac{{x}^{3}}{{({x}^{2}+1)}^{2}}=\frac{Ax+B}{{x}^{2}+1}+\frac{Cx+D}{{({x}^{2}+1)}^{2}}.\]
Finding a common denominator and equating the numerators gives
\[{x}^{3}=(Ax+B)({x}^{2}+1)+Cx+D.\]
Solving, we obtain \(A=1,\) \(B=0,\) \(C=-1,\) and \(D=0.\) Substituting back into the integral, we have
\[\begin{array}{ll}V & =2\pi \int _{0}^{1}\frac{{x}^{3}}{{({x}^{2}+1)}^{2}}dx \\ & =2\pi \int _{0}^{1}(\frac{x}{{x}^{2}+1}-\frac{x}{{({x}^{2}+1)}^{2}})dx \\ & =2\pi (\frac{1}{2}\text{ln}({x}^{2}+1)+\frac{1}{2}\cdot \frac{1}{{x}^{2}+1})|{}_{\begin{array}{l} \\ 0\end{array}}^{\begin{array}{l}1 \\ \end{array}} \\ & =\pi (\text{ln}\ 2-\frac{1}{2}).\end{array}\]
Set up the partial fraction decomposition for \(\int \frac{{x}^{2}+3x+1}{(x+2){(x-3)}^{2}{({x}^{2}+4)}^{2}}dx.\)

Откриј одговор.

\(\frac{{x}^{2}+3x+1}{(x+2){(x-3)}^{2}{({x}^{2}+4)}^{2}}=\frac{A}{x+2}+\frac{B}{x-3}+\frac{C}{{(x-3)}^{2}}+\frac{Dx+E}{{x}^{2}+4}+\frac{Fx+G}{{({x}^{2}+4)}^{2}}\)
\(\frac{1}{(x-3)(x-2)}\)
\(\frac{{x}^{2}+1}{x(x+1)(x+2)}\)

Откриј одговор.

\(-\frac{2}{x+1}+\frac{5}{2(x+2)}+\frac{1}{2x}\)
\(\frac{1}{{x}^{3}-x}\)
\(\frac{3x+1}{{x}^{2}}\)

Откриј одговор.

\(\frac{1}{{x}^{2}}+\frac{3}{x}\)
\(\frac{3{x}^{2}}{{x}^{2}+1}\) (Hint: Use long division first.)
\(\frac{2{x}^{4}}{{x}^{2}-2x}\)

Откриј одговор.

\(2{x}^{2}+4x+8+\frac{16}{x-2}\)
\(\frac{1}{(x-1)({x}^{2}+1)}\)
\(\frac{1}{{x}^{2}(x-1)}\)

Откриј одговор.

\(-\frac{1}{{x}^{2}}-\frac{1}{x}+\frac{1}{x-1}\)
\(\frac{x}{{x}^{2}-4}\)
\(\frac{1}{x(x-1)(x-2)(x-3)}\)

Откриј одговор.

\(-\frac{1}{2(x-2)}+\frac{1}{2(x-1)}-\frac{1}{6x}+\frac{1}{6(x-3)}\)
\(\frac{1}{{x}^{4}-1}=\frac{1}{(x+1)(x-1)({x}^{2}+1)}\)
\(\frac{3{x}^{2}}{{x}^{3}-1}=\frac{3{x}^{2}}{(x-1)({x}^{2}+x+1)}\)

Откриј одговор.

\(\frac{1}{x-1}+\frac{2x+1}{{x}^{2}+x+1}\)
\(\frac{2x}{{(x+2)}^{2}}\)
\(\frac{3{x}^{4}+{x}^{3}+20{x}^{2}+3x+31}{(x+1){({x}^{2}+4)}^{2}}\)

Откриј одговор.

\(\frac{2}{x+1}+\frac{x}{{x}^{2}+4}-\frac{1}{{({x}^{2}+4)}^{2}}\)
\(\int \frac{dx}{(x-3)(x-2)}\)
\(\int \frac{3x}{{x}^{2}+2x-8}dx\)

Откриј одговор.

\(\text{ln}|x-2|+2\ \text{ln}|x+4|+C\)
\(\int \frac{dx}{{x}^{3}-x}\)
\(\int \frac{x}{{x}^{2}-4}dx\)

Откриј одговор.

\(\frac{1}{2}\text{ln}|4-{x}^{2}|+C\)
\(\int \frac{dx}{x(x-1)(x-2)(x-3)}\)
\(\int \frac{2{x}^{2}+4x+22}{{x}^{2}+2x+10}dx\)

Откриј одговор.

\(2(x+\frac{1}{3}\text{arctan}(\frac{1+x}{3}))+C\)
\(\int \frac{dx}{{x}^{2}-5x+6}\)
\(\int \frac{2-x}{{x}^{2}+x}dx\)

Откриј одговор.

\(2\ \text{ln}|x|-3\ \text{ln}|1+x|+C\)
\(\int \frac{2}{{x}^{2}-x-6}dx\)
\(\int \frac{dx}{{x}^{3}-2{x}^{2}-4x+8}\)

Откриј одговор.

\(\frac{1}{16}(\text{-}\ \frac{4}{-2+x}-\text{ln}|-2+x|+\text{ln}|2+x|)+C\)
\(\int \frac{dx}{{x}^{4}-10{x}^{2}+9}\)
\(\int \frac{2}{(x-4)({x}^{2}+2x+6)}dx\)

Откриј одговор.

\(\frac{1}{30}(-2\sqrt{5}\ \text{arctan}[\frac{1+x}{\sqrt{5}}]+2\ \text{ln}|-4+x|-\text{ln}|6+2x+{x}^{2}|)+C\)
\(\int \frac{{x}^{2}}{{x}^{3}-{x}^{2}+4x-4}dx\)
\(\int \frac{{x}^{3}+6{x}^{2}+3x+6}{{x}^{3}+2{x}^{2}}dx\)

Откриј одговор.

\(-\frac{3}{x}+4\ \text{ln}|x+2|+x+C\)

Symbols used here

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Partial Fractions

Integrate a rational function using the method of partial fractions.
Recognize simple linear factors in a rational function.
Recognize repeated linear factors in a rational function.
Recognize quadratic factors in a rational function.
Make sure that
Factor
Assuming that
If

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

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Parts of this page are adapted from OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests