# Integration by Substitution

In this topic we shall see an important method for evaluating many complicated integrals.

Substitution for integrals corresponds to the chain rule for derivatives.

Suppose that $$F\left( u \right)$$ is an antiderivative of $$f\left( u \right):$$

$\int {f\left( u \right)du} = F\left( u \right) + C.$

Assuming that $$u = u\left( x \right)$$ is a differentiable function and using the chain rule, we have

$\frac{d}{{dx}}F\left( {u\left( x \right)} \right) = F^\prime\left( {u\left( x \right)} \right)u^\prime\left( x \right) = f\left( {u\left( x \right)} \right)u^\prime\left( x \right).$

Integrating both sides gives

$\int {f\left( {u\left( x \right)} \right)u^\prime\left( x \right)dx} = F\left( {u\left( x \right)} \right) + C.$

Hence

$\int {{f\left( {u\left( x \right)} \right)}{u^\prime\left( x \right)}dx} = \int {f\left( u \right)du},\;\;\text{where}\;\;u = u\left( x \right).$

This is the substitution rule formula for indefinite integrals.

Note that the integral on the left is expressed in terms of the variable $$x.$$ The integral on the right is in terms of $$u.$$

The substitution method (also called $$u-$$substitution) is used when an integral contains some function and its derivative. In this case, we can set $$u$$ equal to the function and rewrite the integral in terms of the new variable $$u.$$ This makes the integral easier to solve.

Do not forget to express the final answer in terms of the original variable $$x!$$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Compute the integral $\int {{e^{\frac{x}{2}}}dx}.$

### Example 2

Find the integral $\int {{{\left( {3x + 2} \right)}^5}dx}.$

### Example 3

Find the integral $\int {\frac{{dx}}{{\sqrt {1 + 4x} }}}.$

### Example 4

Evaluate the integral $\int {\frac{{xdx}}{{\sqrt {1 + {x^2}} }}}.$

### Example 5

Calculate the integral $\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }}}.$

### Example 6

Evaluate the integral ${\int {{\frac{{{x^2}}}{{{x^3} + 1}}}dx} }$ using an appropriate substitution.

### Example 7

Find the integral $\int {\sqrt[3]{{1 - 3x}}dx}.$

### Example 8

Find the integral $\int {{\frac{{x + 1}}{{{x^2} + 2x - 5}}} dx}.$

### Example 1.

Compute the integral $\int {{e^{\frac{x}{2}}}dx}.$

Solution.

Let $$u = \frac{x}{2}.$$ Then

$du = \frac{{dx}}{2},\;\; \Rightarrow dx = 2du.$

Now we can easily integrate:

$\int {{e^{\frac{x}{2}}}dx} = \int {{e^u} \cdot 2du} = 2\int {{e^u}du} = 2{e^u} + C = 2{e^{\frac{x}{2}}} + C.$

### Example 2.

Find the integral $\int {{{\left( {3x + 2} \right)}^5}dx}.$

Solution.

We make the substitution $$u = 3x + 2.$$ Then

$du = d\left( {3x + 2} \right) = 3dx.$

The differential $$dx$$ is given by

$dx = \frac{{du}}{3}.$

Plug all this in the integral:

$\int {{{\left( {3x + 2} \right)}^5}dx} = \int {{u^5}\frac{{du}}{3}} = \frac{1}{3}\int {{u^5}du} = \frac{1}{3} \cdot \frac{{{u^6}}}{6} + C = \frac{{{u^6}}}{{18}} + C = \frac{{{{\left( {3x + 2} \right)}^6}}}{{18}} + C.$

### Example 3.

Find the integral $\int {\frac{{dx}}{{\sqrt {1 + 4x} }}}.$

Solution.

We can try to use the substitution $$u = 1 + 4x.$$ Hence

$du = d\left( {1 + 4x} \right) = 4dx,$

so

$dx = \frac{{du}}{4}.$

This yields

$\int {\frac{{dx}}{{\sqrt {1 + 4x} }}} = \int {\frac{{\frac{{du}}{4}}}{{\sqrt u }}} = \frac{1}{4}\int {\frac{{du}}{{\sqrt u }}} = \frac{1}{4}\int {{u^{ - \frac{1}{2}}}du} = \frac{1}{4} \cdot \frac{{{u^{\frac{1}{2}}}}}{{\frac{1}{2}}} + C = \frac{1}{4} \cdot 2{u^{\frac{1}{2}}} + C = \frac{{{u^{\frac{1}{2}}}}}{2} + C = \frac{{\sqrt u }}{2} + C = \frac{{\sqrt {1 + 4x} }}{2} + C.$

### Example 4.

Evaluate the integral $\int {\frac{{xdx}}{{\sqrt {1 + {x^2}} }}}.$

Solution.

Let $$u = 1 + {x^2}.$$ Then

$du = d\left( {1 + {x^2}} \right) = 2xdx.$

We see that

$xdx = \frac{{du}}{2}.$

Hence

$\int {\frac{{xdx}}{{\sqrt {1 + {x^2}} }}} = \int {\frac{{\frac{{du}}{2}}}{{\sqrt u }}} = \int {\frac{{du}}{{2\sqrt u }}} = \sqrt u + C = \sqrt {1 + {x^2}} + C.$

### Example 5.

Calculate the integral $\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }}}.$

Solution.

Let $$u = \frac{x}{a}.$$ Then $$x = au,$$ $$dx = adu.$$ Hence, the integral is

$\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }}} = \int {\frac{{adu}}{{\sqrt {{a^2} - {{\left( {au} \right)}^2}} }}} = \int {\frac{{adu}}{{\sqrt {{a^2}\left( {1 - {u^2}} \right)} }}} = \int {\frac{{\cancel{a}du}}{{\cancel{a}\sqrt {1 - {u^2}} }}} = \int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} = \arcsin u + C = \arcsin \frac{x}{a} + C.$

### Example 6.

Evaluate the integral ${\int {{\frac{{{x^2}}}{{{x^3} + 1}}}dx} }$ using an appropriate substitution.

Solution.

We try the substitution $$u = {x^3} + 1.$$

Calculate the differential $$du:$$

$du = d\left( {{x^3} + 1} \right) = 3{x^2}dx.$

We see from the last expression that

${x^2}dx = \frac{{du}}{3},$

so we can rewrite the integral in terms of the new variable $$u:$$

$I = \int {\frac{{{x^2}}}{{{x^3} + 1}}dx} = \int {\frac{{\frac{{du}}{3}}}{u}} = \int {\frac{{du}}{{3u}}} .$

Now we can easily evaluate this integral:

$I = \int {\frac{{du}}{{3u}}} = \frac{1}{3}\int {\frac{{du}}{u}} = {\frac{1}{3}\ln \left| u \right|} + C.$

Express the result in terms of the variable $$x:$$

$I = \frac{1}{3}\ln \left| u \right| + C = \frac{1}{3}\ln \left| {{x^3} + 1} \right| + C.$

### Example 7.

Find the integral $\int {\sqrt[3]{{1 - 3x}}dx}.$

Solution.

We use the substitution $$u = 1 - 3x.$$ Then

$du = d\left( {1 - 3x} \right) = - 3dx$

and

$dx = - \frac{{du}}{3}.$

After substitution we get

$\int {\sqrt[3]{{1 - 3x}}dx} = \int {\sqrt[3]{u}\left( { - \frac{{du}}{3}} \right)} = - \frac{1}{3}\int {\sqrt[3]{u}du} = - \frac{1}{3}\int {{u^{\frac{1}{3}}}du} = - \frac{1}{3} \cdot \frac{{{u^{\frac{4}{3}}}}}{{\frac{4}{3}}} + C = - \frac{1}{3} \cdot \frac{{3{u^{\frac{4}{3}}}}}{4} + C = - \frac{{{u^{\frac{4}{3}}}}}{4} + C = - \frac{{\sqrt[3]{{{u^4}}}}}{4} + C = - \frac{{\sqrt[3]{{{{\left( {1 - 3x} \right)}^4}}}}}{4} + C.$

### Example 8.

Find the integral $\int {{\frac{{x + 1}}{{{x^2} + 2x - 5}}} dx}.$

Solution.

We make the substitution $$u = {x^2} + 2x - 5.$$ Then

$du = 2xdx + 2dx = 2\left( {x + 1} \right)dx$

or

$\left( {x + 1} \right)dx = \frac{{du}}{2}.$

The integral is easy to calculate with the new variable:

$\int {\frac{{x + 1}}{{{x^2} + 2x - 5}}dx} = \int {\frac{{\frac{{du}}{2}}}{u}} = \frac{1}{2}\int {\frac{{du}}{u}} = \frac{1}{2}\ln \left| u \right| + C = \frac{1}{2}\ln \left| {{x^2} + 2x - 5} \right| + C.$

See more problems on Page 2.