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Integrals Involving Exponential and Logarithmic Functions

Integrate functions involving exponential functions.

Integrals of Exponential Functions

The exponential function is perhaps the most efficient function in terms of the operations of calculus. The exponential function, $y={e}^{x},$ is its own derivative and its own integral.

Example

Try it.

Find the antiderivative of the exponential function e^−x.

Solution

Use substitution, setting $u=\text{-}x,$ and then $du=-1dx.$ Multiply the du equation by −1, so you now have $\text{-}du=dx.$ Then,

\[\begin{array}{ll}\int {e}^{\text{-}x}dx & =\text{-}\int {e}^{u}du \\ \\ & =\text{-}{e}^{u}+C \\ & =\text{-}{e}^{\text{-}x}+C.\end{array}\]

A common mistake when dealing with exponential expressions is treating the exponent on e the same way we treat exponents in polynomial expressions. We cannot use the power rule for the exponent on e. This can be especially confusing when we have both exponentials and polynomials in the same expression, as in the previous checkpoint. In these cases, we should always double-check to make sure we’re using the right rules for the functions we’re integrating.

Example

Try it.

Find the antiderivative of the exponential function ${e}^{x}\sqrt{1+{e}^{x}}.$

Solution

First rewrite the problem using a rational exponent:

\[\int {e}^{x}\sqrt{1+{e}^{x}}dx=\int {e}^{x}{(1+{e}^{x})}^{1\text{/}2}dx.\]

Using substitution, choose $u=1+{e}^{x}.$ Then, $du={e}^{x}dx.$ We have ()

\[\int {e}^{x}{(1+{e}^{x})}^{1\text{/}2}dx=\int {u}^{1\text{/}2}du.\]

Then

\[\int {u}^{1\text{/}2}du=\frac{{u}^{3\text{/}2}}{3\text{/}2}+C=\frac{2}{3}{u}^{3\text{/}2}+C=\frac{2}{3}{(1+{e}^{x})}^{3\text{/}2}+C.\]

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

Integrals Involving Logarithmic Functions

Integrating functions of the form $f(x)={x}^{-1}$ result in the absolute value of the natural log function, as shown in the following rule. Integral formulas for other logarithmic functions, such as $f(x)=\text{ln}\ x$ and $f(x)={\text{log}}_{a}x,$ are also included in the rule.

Example

Try it.

Find the antiderivative of the function $\frac{3}{x-10}.$

Solution

First factor the 3 outside the integral symbol. Then use the u⁻¹ rule. Thus,

\[\begin{array}{ll}\int \frac{3}{x-10}dx & =3\int \frac{1}{x-10}dx \\ \\ \\ & =3\int \frac{du}{u} \\ & =3\ \text{ln}|u|+C \\ & =3\ \text{ln}|x-10|+C,x\ne 10.\end{array}\]

See .

Example

Try it.

Find the antiderivative of $\frac{2{x}^{3}+3x}{{x}^{4}+3{x}^{2}}.$

Solution

This can be rewritten as $\int (2{x}^{3}+3x){({x}^{4}+3{x}^{2})}^{-1}dx.$ Use substitution. Let $u={x}^{4}+3{x}^{2},$ then $du=4{x}^{3}+6x.$ Alter du by factoring out the 2. Thus,

\[\begin{array}{lll} \\ du & = & (4{x}^{3}+6x)dx \\ & = & 2(2{x}^{3}+3x)dx \\ \frac{1}{2}\ du & = & (2{x}^{3}+3x)dx.\end{array}\]

Rewrite the integrand in u:

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

Then we have

\[\begin{array}{ll}\frac{1}{2}\int {u}^{-1}du & =\frac{1}{2}\text{ln}|u|+C \\ \\ & =\frac{1}{2}\text{ln}|{x}^{4}+3{x}^{2}|+C.\end{array}\]

Example

Try it.

Find the antiderivative of the log function ${\text{log}}_{2}x.$

Solution

Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Based on this format, we have

\[\int {\text{log}}_{2}xdx=\frac{x}{\text{ln}\ 2}(\text{ln}\ x-1)+C.\]

is a definite integral of a trigonometric function. With trigonometric functions, we often have to apply a trigonometric property or an identity before we can move forward. Finding the right form of the integrand is usually the key to a smooth integration.

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

Key Concepts

Exponential and logarithmic functions arise in many real-world applications, especially those involving growth and decay.
Substitution is often used to evaluate integrals involving exponential functions or logarithms.

Key Equations

Integrals of Exponential Functions	$\int {e}^{x}dx={e}^{x}+C$ $\int {a}^{x}dx=\frac{{a}^{x}}{\text{ln}\ a}+C$
Integration Formulas Involving Logarithmic Functions	$\int {x}^{-1}dx=\text{ln}\|x\|+C$ $\int \text{ln}\ x\ dx=x\ \text{ln}\ x-x+C=x(\text{ln}\ x-1)+C$ $\int {\text{log}}_{a}\ x\ dx=\frac{x}{\text{ln}\ a}(\text{ln}\ x-1)+C$

Integrals Involving Exponential and Logarithmic Functions

In the following exercises, compute each indefinite integral.

In the following exercises, find each indefinite integral by using appropriate substitutions.

In the following exercises, verify by differentiation that $\int \text{ln}\ x\ dx=x(\text{ln}\ x-1)+C,$ then use appropriate changes of variables to compute the integral.

In the following exercises, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms.

In the following exercises, evaluate the definite integral.

In the following exercises, integrate using the indicated substitution.

In the following exercises, does the right-endpoint approximation overestimate or underestimate the exact area? Calculate the right endpoint estimate R₅₀ and solve for the exact area.

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

Integrals of Exponential Functions

The exponential function is perhaps the most efficient function in terms of the operations of calculus. The exponential function, $y={e}^{x},$ is its own derivative and its own integral.

Example

Try it.

Find the antiderivative of the exponential function e^−x.

Solution

Use substitution, setting $u=\text{-}x,$ and then $du=-1dx.$ Multiply the du equation by −1, so you now have $\text{-}du=dx.$ Then,

\[\begin{array}{ll}\int {e}^{\text{-}x}dx & =\text{-}\int {e}^{u}du \\ \\ & =\text{-}{e}^{u}+C \\ & =\text{-}{e}^{\text{-}x}+C.\end{array}\]

Example

Try it.

Find the antiderivative of the exponential function ${e}^{x}\sqrt{1+{e}^{x}}.$

Solution

First rewrite the problem using a rational exponent:

\[\int {e}^{x}\sqrt{1+{e}^{x}}dx=\int {e}^{x}{(1+{e}^{x})}^{1\text{/}2}dx.\]

Using substitution, choose $u=1+{e}^{x}.$ Then, $du={e}^{x}dx.$ We have ()

\[\int {e}^{x}{(1+{e}^{x})}^{1\text{/}2}dx=\int {u}^{1\text{/}2}du.\]

Then

\[\int {u}^{1\text{/}2}du=\frac{{u}^{3\text{/}2}}{3\text{/}2}+C=\frac{2}{3}{u}^{3\text{/}2}+C=\frac{2}{3}{(1+{e}^{x})}^{3\text{/}2}+C.\]

Condensed — the full section is in OpenStax Calculus Volume 1.

Integrals Involving Logarithmic Functions

Example

Try it.

Find the antiderivative of the function $\frac{3}{x-10}.$

Solution

First factor the 3 outside the integral symbol. Then use the u⁻¹ rule. Thus,

\[\begin{array}{ll}\int \frac{3}{x-10}dx & =3\int \frac{1}{x-10}dx \\ \\ \\ & =3\int \frac{du}{u} \\ & =3\ \text{ln}|u|+C \\ & =3\ \text{ln}|x-10|+C,x\ne 10.\end{array}\]

See .

Example

Try it.

Find the antiderivative of $\frac{2{x}^{3}+3x}{{x}^{4}+3{x}^{2}}.$

Solution

This can be rewritten as $\int (2{x}^{3}+3x){({x}^{4}+3{x}^{2})}^{-1}dx.$ Use substitution. Let $u={x}^{4}+3{x}^{2},$ then $du=4{x}^{3}+6x.$ Alter du by factoring out the 2. Thus,

\[\begin{array}{lll} \\ du & = & (4{x}^{3}+6x)dx \\ & = & 2(2{x}^{3}+3x)dx \\ \frac{1}{2}\ du & = & (2{x}^{3}+3x)dx.\end{array}\]

Rewrite the integrand in u:

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

Then we have

\[\begin{array}{ll}\frac{1}{2}\int {u}^{-1}du & =\frac{1}{2}\text{ln}|u|+C \\ \\ & =\frac{1}{2}\text{ln}|{x}^{4}+3{x}^{2}|+C.\end{array}\]

Example

Try it.

Find the antiderivative of the log function ${\text{log}}_{2}x.$

Solution

Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Based on this format, we have

\[\int {\text{log}}_{2}xdx=\frac{x}{\text{ln}\ 2}(\text{ln}\ x-1)+C.\]

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

Exponential and logarithmic functions arise in many real-world applications, especially those involving growth and decay.
Substitution is often used to evaluate integrals involving exponential functions or logarithms.

Key Equations

Integrals of Exponential Functions	$\int {e}^{x}dx={e}^{x}+C$ $\int {a}^{x}dx=\frac{{a}^{x}}{\text{ln}\ a}+C$
Integration Formulas Involving Logarithmic Functions	$\int {x}^{-1}dx=\text{ln}\|x\|+C$ $\int \text{ln}\ x\ dx=x\ \text{ln}\ x-x+C=x(\text{ln}\ x-1)+C$ $\int {\text{log}}_{a}\ x\ dx=\frac{x}{\text{ln}\ a}(\text{ln}\ x-1)+C$

Integrals Involving Exponential and Logarithmic Functions

In the following exercises, compute each indefinite integral.

In the following exercises, find each indefinite integral by using appropriate substitutions.

In the following exercises, verify by differentiation that $\int \text{ln}\ x\ dx=x(\text{ln}\ x-1)+C,$ then use appropriate changes of variables to compute the integral.

In the following exercises, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms.

In the following exercises, evaluate the definite integral.

In the following exercises, integrate using the indicated substitution.

In the following exercises, does the right-endpoint approximation overestimate or underestimate the exact area? Calculate the right endpoint estimate R₅₀ and solve for the exact area.

Condensed — the full section is in OpenStax Calculus Volume 1.

Practice (40)

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

Find the antiderivative of the exponential function e^−x.

Αποκάλυψέ την.

Use substitution, setting $u=\text{-}x,$ and then $du=-1dx.$ Multiply the du equation by −1, so you now have $\text{-}du=dx.$ Then,
\[\begin{array}{ll}\int {e}^{\text{-}x}dx & =\text{-}\int {e}^{u}du \\ \\ & =\text{-}{e}^{u}+C \\ & =\text{-}{e}^{\text{-}x}+C.\end{array}\]
Find the antiderivative of the function using substitution: ${x}^{2}{e}^{-2{x}^{3}}.$

Αποκάλυψέ την.

$\int {x}^{2}{e}^{-2{x}^{3}}dx=-\frac{1}{6}{e}^{-2{x}^{3}}+C$
Find the antiderivative of the exponential function ${e}^{x}\sqrt{1+{e}^{x}}.$

Αποκάλυψέ την.

First rewrite the problem using a rational exponent:
\[\int {e}^{x}\sqrt{1+{e}^{x}}dx=\int {e}^{x}{(1+{e}^{x})}^{1\text{/}2}dx.\]
Using substitution, choose $u=1+{e}^{x}.$ Then, $du={e}^{x}dx.$ We have ()
\[\int {e}^{x}{(1+{e}^{x})}^{1\text{/}2}dx=\int {u}^{1\text{/}2}du.\]
Then
\[\int {u}^{1\text{/}2}du=\frac{{u}^{3\text{/}2}}{3\text{/}2}+C=\frac{2}{3}{u}^{3\text{/}2}+C=\frac{2}{3}{(1+{e}^{x})}^{3\text{/}2}+C.\]
Find the antiderivative of ${e}^{x}{(3{e}^{x}-2)}^{2}.$

Αποκάλυψέ την.

$\int {e}^{x}{(3{e}^{x}-2)}^{2}dx=\frac{1}{9}{(3{e}^{x}-2)}^{3}$
Use substitution to evaluate the indefinite integral $\int 3{x}^{2}{e}^{2{x}^{3}}dx.$

Αποκάλυψέ την.

Here we choose to let u equal the expression in the exponent on e. Let $u=2{x}^{3}$ and $du=6{x}^{2}dx..$ Again, du is off by a constant multiplier; the original function contains a factor of 3x², not 6x². Multiply both sides of the equation by $\frac{1}{2}$ so that the integrand in u equals the integrand in x. Thus,
\[\int 3{x}^{2}{e}^{2{x}^{3}}dx=\frac{1}{2}\int {e}^{u}du.\]
Integrate the expression in u and then substitute the original expression in x back into the u integral:
\[\frac{1}{2}\int {e}^{u}du=\frac{1}{2}{e}^{u}+C=\frac{1}{2}{e}^{2{x}^{3}}+C.\]
Evaluate the indefinite integral $\int 2{x}^{3}{e}^{{x}^{4}}dx.$

Αποκάλυψέ την.

$\int 2{x}^{3}{e}^{{x}^{4}}dx=\frac{1}{2}{e}^{{x}^{4}}$
Find the price–demand equation for a particular brand of toothpaste at a supermarket chain when the demand is 50 tubes per week at $2.35 per tube, given that the marginal price—demand function, ${p}^{'}\text{(}x),$ for x number of tubes per week, is given as
\[p'(x)=-0.015{e}^{-0.01x}.\]
If the supermarket chain sells 100 tubes per week, what price should it set?

Αποκάλυψέ την.

To find the price–demand equation, integrate the marginal price–demand function. First find the antiderivative, then look at the particulars. Thus,
\[\begin{array}{ll} \\ \\ p(x) & =\int -0.015{e}^{-0.01x}dx \\ & =-0.015\int {e}^{-0.01x}dx.\end{array}\]
Using substitution, let $u=-0.01x$ and $du=-0.01dx.$ Then, divide both sides of the du equation by −0.01. This gives
\[\begin{array}{ll}\frac{-0.015}{-0.01}\int {e}^{u}du & =1.5\int {e}^{u}du \\ \\ & =1.5{e}^{u}+C \\ & =1.5{e}^{-0.01x}+C.\end{array}\]
The next step is to solve for C. We know that when the price is $2.35 per tube, the demand is 50 tubes per week. This means
\[\begin{array}{ll} \\ \\ p(50) & =1.5{e}^{-0.01(50)}+C \\ & =2.35.\end{array}\]
Now, just solve for C:
\[\begin{array}{ll} \\ C & =2.35-1.5{e}^{-0.5} \\ & =2.35-0.91 \\ & =1.44.\end{array}\]
Thus,
\[p(x)=1.5{e}^{-0.01x}+1.44.\]
If the supermarket sells 100 tubes of toothpaste per week, the price would be
\[p(100)=1.5{e}^{-0.01(100)}+1.44=1.5{e}^{-1}+1.44\approx 1.99.\]
The supermarket should charge $1.99 per tube if it is selling 100 tubes per week.
Evaluate the definite integral ${\int }_{1}^{2}{e}^{1-x}dx.$

Αποκάλυψέ την.

Again, substitution is the method to use. Let $u=1-x,$ so $du=-1dx$ or $\text{-}du=dx.$ Then $\int {e}^{1-x}dx=\text{-}\int {e}^{u}du.$ Next, change the limits of integration. Using the equation $u=1-x,$ we have
\[\begin{array}{l}u=1-(1)=0 \\ u=1-(2)=-1.\end{array}\]
The integral then becomes
\[\begin{array}{ll}{\int }_{1}^{2}{e}^{1-x}dx & =\text{-}{\int }_{0}^{-1}{e}^{u}du \\ \\ \\ & ={\int }_{-1}^{0}{e}^{u}du \\ & ={{e}^{u}|}_{-1}^{0} \\ & ={e}^{0}-({e}^{-1}) \\ & =\text{-}{e}^{-1}+1.\end{array}\]
See .
Evaluate ${\int }_{0}^{2}{e}^{2x}dx.$

Αποκάλυψέ την.

$\frac{1}{2}{\int }_{0}^{4}{e}^{u}du=\frac{1}{2}({e}^{4}-1)$
Suppose the rate of growth of bacteria in a Petri dish is given by $q(t)={3}^{t},$ where t is given in hours and $q(t)$ is given in thousands of bacteria per hour. If a culture starts with 10,000 bacteria, find a function $Q(t)$ that gives the number of bacteria in the Petri dish at any time t. How many bacteria are in the dish after 2 hours?

Αποκάλυψέ την.

We have
\[Q(t)=\int {3}^{t}dt=\frac{{3}^{t}}{\text{ln}\ 3}+C.\]
Then, at $t=0$ we have $Q(0)=10=\frac{1}{\text{ln}\ 3}+C,$ so $C\approx 9.090$ and we get
\[Q(t)=\frac{{3}^{t}}{\text{ln}\ 3}+9.090.\]
At time $t=2,$ we have
\[Q(2)=\frac{{3}^{2}}{\text{ln}\ 3}+9.090\]\[=17.282.\]
After 2 hours, there are 17,282 bacteria in the dish.
From , suppose the bacteria grow at a rate of $q(t)={2}^{t}.$ Assume the culture still starts with 10,000 bacteria. Find $Q(t).$ How many bacteria are in the dish after 3 hours?

Αποκάλυψέ την.

$Q(t)=\frac{{2}^{t}}{\text{ln}\ 2}+8.557.$ There are 20,099 bacteria in the dish after 3 hours.
Suppose a population of fruit flies increases at a rate of $g(t)=2{e}^{0.02t},$ in flies per day. If the initial population of fruit flies is 100 flies, how many flies are in the population after 10 days?

Αποκάλυψέ την.

Let $G(t)$ represent the number of flies in the population at time t. Applying the net change theorem, we have
\[\begin{array}{ll} \\ \\ G(10) & =G(0)+{\int }_{0}^{10}2{e}^{0.02t}dt \\ & =100+{[\frac{2}{0.02}{e}^{0.02t}]|}_{0}^{10} \\ & =100+{[100{e}^{0.02t}]|}_{0}^{10} \\ & =100+100{e}^{0.2}-100 \\ & \approx 122.\end{array}\]
There are 122 flies in the population after 10 days.
Suppose the rate of growth of the fly population is given by $g(t)={e}^{0.01t},$ and the initial fly population is 100 flies. How many flies are in the population after 15 days?

Αποκάλυψέ την.

There are 116 flies.
Evaluate the definite integral using substitution: ${\int }_{1}^{2}\frac{{e}^{1\text{/}x}}{{x}^{2}}dx.$

Αποκάλυψέ την.

This problem requires some rewriting to simplify applying the properties. First, rewrite the exponent on e as a power of x, then bring the x² in the denominator up to the numerator using a negative exponent. We have
\[{\int }_{1}^{2}\frac{{e}^{1\text{/}x}}{{x}^{2}}dx={\int }_{1}^{2}{e}^{{x}^{-1}}{x}^{-2}dx.\]
Let $u={x}^{-1},$ the exponent on e. Then
\[\begin{array}{ll}du & =\text{-}{x}^{-2}dx \\ -du & ={x}^{-2}dx.\end{array}\]
Bringing the negative sign outside the integral sign, the problem now reads
\[\text{-}\int {e}^{u}du.\]
Next, change the limits of integration:
\[\begin{array}{l} \\ \\ u={(1)}^{-1}=1 \\ u={(2)}^{-1}=\frac{1}{2}.\end{array}\]
Notice that now the limits begin with the larger number, meaning we must multiply by −1 and interchange the limits. Thus,
\[\begin{array}{ll} \\ \\ \\ \text{-}{\int }_{1}^{1\text{/}2}{e}^{u}du & ={\int }_{1\text{/}2}^{1}{e}^{u}du \\ & ={e}^{u}{|}_{1\text{/}2}^{1} \\ & =e-{e}^{1\text{/}2} \\ & =e-\sqrt{e}.\end{array}\]
Evaluate the definite integral using substitution: ${\int }_{1}^{2}\frac{1}{{x}^{3}}{e}^{4{x}^{-2}}dx.$

Αποκάλυψέ την.

${\int }_{1}^{2}\frac{1}{{x}^{3}}{e}^{4{x}^{-2}}dx=\frac{1}{8}[{e}^{4}-e]$
Find the antiderivative of the function $\frac{3}{x-10}.$

Αποκάλυψέ την.

First factor the 3 outside the integral symbol. Then use the u⁻¹ rule. Thus,
\[\begin{array}{ll}\int \frac{3}{x-10}dx & =3\int \frac{1}{x-10}dx \\ \\ \\ & =3\int \frac{du}{u} \\ & =3\ \text{ln}|u|+C \\ & =3\ \text{ln}|x-10|+C,x\ne 10.\end{array}\]
See .
Find the antiderivative of $\frac{1}{x+2}.$

Αποκάλυψέ την.

$\text{ln}|x+2|+C$
Find the antiderivative of $\frac{2{x}^{3}+3x}{{x}^{4}+3{x}^{2}}.$

Αποκάλυψέ την.

This can be rewritten as $\int (2{x}^{3}+3x){({x}^{4}+3{x}^{2})}^{-1}dx.$ Use substitution. Let $u={x}^{4}+3{x}^{2},$ then $du=4{x}^{3}+6x.$ Alter du by factoring out the 2. Thus,
\[\begin{array}{lll} \\ du & = & (4{x}^{3}+6x)dx \\ & = & 2(2{x}^{3}+3x)dx \\ \frac{1}{2}\ du & = & (2{x}^{3}+3x)dx.\end{array}\]
Rewrite the integrand in u:
\[\int (2{x}^{3}+3x){({x}^{4}+3{x}^{2})}^{-1}dx=\frac{1}{2}\int {u}^{-1}du.\]
Then we have
\[\begin{array}{ll}\frac{1}{2}\int {u}^{-1}du & =\frac{1}{2}\text{ln}|u|+C \\ \\ & =\frac{1}{2}\text{ln}|{x}^{4}+3{x}^{2}|+C.\end{array}\]
Find the antiderivative of the log function ${\text{log}}_{2}x.$

Αποκάλυψέ την.

Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Based on this format, we have
\[\int {\text{log}}_{2}xdx=\frac{x}{\text{ln}\ 2}(\text{ln}\ x-1)+C.\]
Find the antiderivative of ${\text{log}}_{3}x.$

Αποκάλυψέ την.

$\frac{x}{\text{ln}\ 3}(\text{ln}\ x-1)+C$
Find the definite integral of ${\int }_{0}^{\pi \text{/}2}\frac{\text{sin}\ x}{1+\text{cos}\ x}dx.$

Αποκάλυψέ την.

We need substitution to evaluate this problem. Let $u=1+\text{cos}\ x,,$ so $du=\text{-}\text{sin}\ x\ dx.$ Rewrite the integral in terms of u, changing the limits of integration as well. Thus,
\[\begin{array}{l}u=1+\text{cos}(0)=2 \\ u=1+\text{cos}(\frac{\pi }{2})=1.\end{array}\]
Then
\[\begin{array}{ll}{\int }_{0}^{\pi \text{/}2}\frac{\text{sin}\ x}{1+\text{cos}\ x} & =\text{-}{\int }_{2}^{1}{u}^{-1}du \\ \\ \\ & ={\int }_{1}^{2}{u}^{-1}du \\ & ={\text{ln}|u||}_{1}^{2} \\ & =[\text{ln}\ 2-\text{ln}\ 1] \\ & =\text{ln}\ 2.\end{array}\]
$\int {e}^{2x}dx$
$\int {e}^{-3x}dx$

Αποκάλυψέ την.

$\frac{-1}{3}{e}^{-3x}+C$
$\int {2}^{x}dx$
$\int {3}^{\text{-}x}dx$

Αποκάλυψέ την.

$-\frac{{3}^{\text{-}x}}{\text{ln}\ 3}+C$
$\int \frac{1}{2x}dx$
$\int \frac{2}{x}dx$

Αποκάλυψέ την.

$2\ln \left|x\right|+C\text{or}\ln \left({x}^{2}\right)\text{+C}$
$\int \frac{1}{{x}^{2}}dx$
$\int \frac{1}{\sqrt{x}}dx$

Αποκάλυψέ την.

$2\sqrt{x}+C$
$\int \frac{\text{ln}\ x}{x}dx$
$\int \frac{dx}{x{(\text{ln}\ x)}^{2}}$

Αποκάλυψέ την.

$-\frac{1}{\text{ln}\ x}+C$
$\int \frac{dx}{x\ \text{ln}\ x}\ (x>1)$
$\int \frac{dx}{x\ \text{ln}\ x\ \text{ln}(\text{ln}\ x)}$

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$\text{ln}(\text{ln}(\text{ln}\ x))+C$
$\int \text{tan}\ \theta \ d\theta$
$\int \frac{\text{cos}\ x-x\ \text{sin}\ x}{x\ \text{cos}\ x}dx$

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$\text{ln}(x\ \text{cos}\ x)+C$
$\int \frac{\text{ln}(\text{sin}\ x)}{\text{tan}\ x}dx$
$\int \text{ln}(\text{cos}\ x)\text{tan}\ xdx$

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$-\frac{1}{2}{(\text{ln}(\text{cos}(x)))}^{2}+C$
$\int x{e}^{\text{-}{x}^{2}}dx$
$\int {x}^{2}{e}^{\text{-}{x}^{3}}dx$

Αποκάλυψέ την.

$\frac{\text{-}{e}^{\text{-}{x}^{3}}}{3}+C$
$\int {e}^{\text{sin}\ x}\text{cos}\ xdx$

Symbols used here

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

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

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

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

derivative
Instantaneous rate of change; slope of the graph.

\neq

not equal
The two sides are different.

\pi

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

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: Integrals Involving Exponential and Logarithmic Functions

Integrate functions involving exponential functions.
Integrate functions involving logarithmic functions.
Exponential and logarithmic functions arise in many real-world applications, especially those involving growth and decay.
Substitution is often used to evaluate integrals involving exponential functions or logarithms.

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.

Δοκίμασε μόνος σου.

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Περισσότερα σε Calculus

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

Integrals of Exponential Functions	\(\int {e}^{x}dx={e}^{x}+C\) \(\int {a}^{x}dx=\frac{{a}^{x}}{\text{ln}\ a}+C\)
Integration Formulas Involving Logarithmic Functions	\(\int {x}^{-1}dx=\text{ln}\|x\|+C\) \(\int \text{ln}\ x\ dx=x\ \text{ln}\ x-x+C=x(\text{ln}\ x-1)+C\) \(\int {\text{log}}_{a}\ x\ dx=\frac{x}{\text{ln}\ a}(\text{ln}\ x-1)+C\)