Integration by substitution

Integration by substitution simplifies a composite integrand by replacing an inner expression with a new variable. It is the integration counterpart of the chain rule.

The key recognition is

$$\int f(g(x))g'(x)\,dx=F(g(x))+C,$$

where $F'(u)=f(u)$. In words: look for an inner function and its derivative, perhaps multiplied by a constant.

Prerequisites

You should be able to:

• apply the chain rule
• differentiate powers, exponentials, logarithms and trigonometric functions
• use the standard results in integration basics
• rearrange expressions and change the limits of a definite integral

Why substitution works

Suppose $u=g(x)$. Then

$$\frac{du}{dx}=g'(x),$$

so the differential notation $du=g'(x)\,dx$ records the factor needed when changing variable. Therefore

$$\int f(g(x))g'(x)\,dx =\int f(u)\,du =F(u)+C =F(g(x))+C.$$

This is exactly the chain rule read backwards. For example,

$$\frac{d}{dx}(x^2+1)^5=10x(x^2+1)^4.$$

Hence

$$\int 10x(x^2+1)^4\,dx=(x^2+1)^5+C.$$

The symbols $dx$ and $du$ are not decoration. They identify the variable with respect to which the integration is taking place.

The substitution method

For an indefinite integral:

1. Choose $u$ to be an expression that simplifies the integrand.
2. Differentiate it to obtain $du$ in terms of $dx$.
3. Rewrite the entire integral in terms of $u$ and $du$.
4. Integrate with respect to $u$.
5. Substitute back to $x$ and add $C$.
6. Differentiate the answer to check it.

A good substitution usually removes a repeated inner expression, a root, a denominator, or a complicated power.

Direct reverse chain rule

Worked example 1: a power of a linear expression

Find

$$\int (3x-2)^7\,dx.$$

Let

$$u=3x-2.$$

Then $du=3\,dx$, so $dx=\frac13\,du$. Replace both the bracket and $dx$:

$$\begin{aligned} \int(3x-2)^7\,dx &=\int u^7\left(\frac13\,du\right)\\ &=\frac13\cdot\frac{u^8}{8}+C\\ &=\boxed{\frac{(3x-2)^8}{24}+C}. \end{aligned}$$

Check:

$$\frac{d}{dx}\left(\frac{(3x-2)^8}{24}\right) =\frac8{24}(3x-2)^7(3) =(3x-2)^7.$$

The factor $\frac13$ compensates for the derivative of $3x-2$.

Worked example 2: the derivative is present up to a constant

Find

$$\int x(x^2+4)^6\,dx.$$

Choose $u=x^2+4$. Since $du=2x\,dx$,

$$x\,dx=\frac12\,du.$$

Therefore

$$\begin{aligned} \int x(x^2+4)^6\,dx &=\frac12\int u^6\,du\\ &=\frac{u^7}{14}+C\\ &=\boxed{\frac{(x^2+4)^7}{14}+C}. \end{aligned}$$

It is not necessary for $2x$ to appear exactly. A non-zero constant multiple can be adjusted outside the integral.

Logarithmic forms

The pattern

$$\boxed{\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C}$$

is substitution with $u=f(x)$.

Worked example 3: numerator related to the denominator

Find

$$\int\frac{6x-1}{3x^2-x+5}\,dx.$$

Let

$$u=3x^2-x+5, \qquad du=(6x-1)\,dx.$$

The integral becomes

$$\int\frac1u\,du =\ln|u|+C =\boxed{\ln|3x^2-x+5|+C}.$$

Here $3x^2-x+5>0$ for every real $x$, so $\ln(3x^2-x+5)+C$ is also valid. Keeping the absolute value is the safe general form.

Worked example 4: split the numerator first

Find

$$\int\frac{x+3}{x^2+4x+7}\,dx.$$

The derivative of the denominator is $2x+4$, but the numerator is not a constant multiple of it. Rewrite

$$x+3=\frac12(2x+4)+1.$$

Then

$$\begin{aligned} \int\frac{x+3}{x^2+4x+7}\,dx &=\frac12\int\frac{2x+4}{x^2+4x+7}\,dx +\int\frac1{x^2+4x+7}\,dx. \end{aligned}$$

The first integral is logarithmic. For the second, complete the square:

$$x^2+4x+7=(x+2)^2+3.$$

Using $\int\frac1{z^2+a^2}\,dz=\frac1a\arctan\frac za+C$ gives

$$\boxed{ \frac12\ln(x^2+4x+7) +\frac1{\sqrt3}\arctan\left(\frac{x+2}{\sqrt3}\right)+C }.$$

This example shows an important limit of pattern matching: substitution may solve only part of an integral.

Exponential and trigonometric composites

Worked example 5: exponential composite

Evaluate

$$\int 4x e^{2x^2+1}\,dx.$$

Let $u=2x^2+1$. Then $du=4x\,dx$, so

$$\int 4x e^{2x^2+1}\,dx =\int e^u\,du =e^u+C =\boxed{e^{2x^2+1}+C}.$$

Worked example 6: trigonometric composite

Find

$$\int \sin^3x\cos x\,dx.$$

Because $\frac{d}{dx}(\sin x)=\cos x$, choose $u=\sin x$. Then $du=\cos x\,dx$:

$$\begin{aligned} \int \sin^3x\cos x\,dx &=\int u^3\,du\\ &=\frac{u^4}{4}+C\\ &=\boxed{\frac14\sin^4x+C}. \end{aligned}$$

The notation $\sin^3x$ means $(\sin x)^3$.

Worked example 7: a reciprocal trigonometric result

Find

$$\int\frac{\cos x}{2+\sin x}\,dx.$$

Let $u=2+\sin x$. Then $du=\cos x\,dx$, giving

$$\boxed{\int\frac{\cos x}{2+\sin x}\,dx =\ln|2+\sin x|+C}.$$

Since $2+\sin x\geq1$, the expression inside the logarithm is always positive.

Substitutions that change the form completely

Sometimes the useful substitution is supplied in a question. You must transform every occurrence of $x$ and $dx$.

Worked example 8: removing a square root

Use $u=\sqrt{x+1}$ to find

$$\int\frac{x}{\sqrt{x+1}}\,dx.$$

From $u=\sqrt{x+1}$,

$$u^2=x+1, \qquad x=u^2-1, \qquad dx=2u\,du.$$

Now transform the whole integrand:

$$\begin{aligned} \int\frac{x}{\sqrt{x+1}}\,dx &=\int\frac{u^2-1}{u}(2u)\,du\\ &=\int(2u^2-2)\,du\\ &=\frac23u^3-2u+C. \end{aligned}$$

Substitute back:

$$\boxed{\frac23(x+1)^{3/2}-2\sqrt{x+1}+C}.$$

An equivalent simplified answer is $\frac23(x-2)\sqrt{x+1}+C$. Different looking antiderivatives are acceptable if they differ only by a constant.

Definite integrals and changing limits

For a definite integral, change the limits from $x$ values to $u$ values as soon as you substitute:

$$\int_a^b f(g(x))g'(x)\,dx =\int_{g(a)}^{g(b)}f(u)\,du.$$

Once the limits are in $u$, do not substitute back to $x$. Alternatively, find an antiderivative in $x$ and retain the original limits. Do not mix the two approaches.

Worked example 9: changing the limits

Evaluate

$$\int_0^1 2x(x^2+3)^4\,dx.$$

Let $u=x^2+3$, so $du=2x\,dx$. Transform the limits:

$$x=0\Rightarrow u=3, \qquad x=1\Rightarrow u=4.$$

Therefore

$$\begin{aligned} \int_0^1 2x(x^2+3)^4\,dx &=\int_3^4u^4\,du\\ &=\left[\frac{u^5}{5}\right]_3^4\\ &=\boxed{\frac{4^5-3^5}{5}}\\ &=\boxed{\frac{781}{5}}. \end{aligned}$$

Worked example 10: a decreasing substitution

Evaluate

$$\int_0^1\frac{x}{(4-x^2)^2}\,dx.$$

Let $u=4-x^2$. Then $du=-2x\,dx$, so $x\,dx=-\frac12du$. The new limits are

$$x=0\Rightarrow u=4, \qquad x=1\Rightarrow u=3.$$

Thus

$$\begin{aligned} \int_0^1\frac{x}{(4-x^2)^2}\,dx &=-\frac12\int_4^3u^{-2}\,du\\ &=\frac12\int_3^4u^{-2}\,du\\ &=\frac12\left[-u^{-1}\right]_3^4\\ &=\boxed{\frac1{24}}. \end{aligned}$$

The reversed $u$ limits are correct because $u=4-x^2$ decreases as $x$ increases. Reversing the limits introduces a second minus sign.

Choosing an effective substitution

Ask these questions in order:

1. Is there a composite expression whose derivative also appears?
2. Does a denominator have its derivative in the numerator?
3. Would replacing a repeated bracket, root, exponential or trigonometric expression remove most occurrences of $x$?
4. Can algebraic simplification make the pattern visible first?

Integrand	Useful choice	Reason
$x(1+x^2)^8$	$u=1+x^2$	$du=2x\,dx$
$\dfrac{e^x}{1+e^x}$	$u=1+e^x$	numerator is $u'$
$\cos(5x-1)$	$u=5x-1$	linear inner function
$\dfrac{(\ln x)^4}{x}$	$u=\ln x$	$du=\dfrac1x\,dx$
$\dfrac{x}{\sqrt{x^2+9}}$	$u=x^2+9$	removes the root and $x\,dx$

Substitution is not automatic whenever brackets appear. For example,

$$\int x e^x\,dx$$

has no single inner function whose derivative accounts for the other factor. It is naturally handled by integration by parts. A rational function may instead require partial fractions.

Common misconceptions

Forgetting to transform $dx$

If $u=x^2+1$, writing

$$\int x(x^2+1)^5\,dx=\int xu^5\,du$$

is invalid. The variables have been mixed. Use $du=2x\,dx$, so $x\,dx=\frac12du$.

Missing the constant factor

$$\int\cos(4x)\,dx\ne\sin(4x)+C.$$

Differentiating the right side gives $4\cos(4x)$. The correct result is

$$\int\cos(4x)\,dx=\frac14\sin(4x)+C.$$

Leaving both $x$ and $u$

Before integrating, the new integral should contain only $u$ and $du$. If an $x$ remains, either express it in terms of $u$ or choose a better substitution.

Using old limits with a new variable

After changing from $x$ to $u$, the limits must also be $u$ values. Limits $0$ and $1$ do not retain their meaning merely because they are numbers.

Adding $C$ to a definite integral

Add $C$ to an indefinite integral. A definite integral is a number, and any constant cancels when the limits are applied.

Self-check

1. Linear inner function

Find

$$\int(5x+2)^{-3}\,dx.$$

Answer

Let $u=5x+2$, so $dx=\frac15du$.

$$\int(5x+2)^{-3}\,dx =\frac15\int u^{-3}\,du =-\frac1{10}u^{-2}+C =\boxed{-\frac1{10(5x+2)^2}+C}.$$

2. Logarithmic substitution

Find

$$\int\frac{3x^2}{x^3-7}\,dx.$$

Answer

Take $u=x^3-7$, so $du=3x^2\,dx$:

$$\boxed{\int\frac{3x^2}{x^3-7}\,dx=\ln|x^3-7|+C}.$$

The absolute value is needed because $x^3-7$ can be negative.

3. Exponential substitution

Find

$$\int\frac{e^{2x}}{3+e^{2x}}\,dx.$$

Answer

Let $u=3+e^{2x}$. Then $du=2e^{2x}\,dx$, so

$$\boxed{\int\frac{e^{2x}}{3+e^{2x}}\,dx =\frac12\ln(3+e^{2x})+C}.$$

4. Definite integral

Evaluate

$$\int_0^{\pi/4}\tan x\,dx.$$

Answer

Write $\tan x=\frac{\sin x}{\cos x}$ and let $u=\cos x$. Then $du=-\sin x\,dx$. The limits become $u=1$ and $u=\frac1{\sqrt2}$:

$$\begin{aligned} \int_0^{\pi/4}\tan x\,dx &=-\int_1^{1/\sqrt2}\frac1u\,du\\ &=-[\ln u]_1^{1/\sqrt2}\\ &=\boxed{\frac12\ln2}. \end{aligned}$$

5. Diagnose the error

A student claims

$$\int 2x\sqrt{x^2+5}\,dx=\frac23(x^2+5)^{3/2}+C.$$

Is the answer correct?

Answer

Yes. With $u=x^2+5$, $du=2x\,dx$, so

$$\int2x\sqrt{x^2+5}\,dx =\int u^{1/2}\,du =\frac23u^{3/2}+C.$$

Do not reject a correct answer merely because the chain rule would produce an extra factor: here the factor $2x$ is already present.

Exam strategy and next steps

• State the substitution clearly, including $du$ or $\frac{du}{dx}$.
• Show how any constant factor is handled.
• For definite integrals, display the transformed limits before integrating.
• Check an indefinite answer by differentiation.
• Estimate the sign and size of a definite answer when possible.

Next, learn how reversing the product rule leads to integration by parts, and how algebraic decomposition prepares rational functions for integration using partial fractions. Return to definite integrals and areas when substitution is used inside area or accumulation problems.