The reverse chain rule

Integrating functions of the form f'(x)[f(x)]^n with the reverse chain rule, adjusting for a missing constant factor, and evaluating definite integrals this way.

The rule

The chain rule gives $\dfrac{d}{dx}\dfrac{[f(x)]^{n+1}}{n+1} = f'(x)\,[f(x)]^n$. Reading this backwards gives the rule.

Reverse chain rule

$$\int f'(x)\,[f(x)]^n\,dx = \frac{[f(x)]^{n+1}}{n + 1} + C, \quad n \ne -1$$

Example 1●

Find $\displaystyle\int 2x(x^2 + 1)^3\,dx$.

$f(x) = x^2 + 1$ and $f'(x) = 2x$ is present.

$\dfrac{(x^2 + 1)^4}{4} + C$

Adjusting the constant

If the derivative is there apart from a constant factor, multiply and divide by that factor.

Example 2●●

Find $\displaystyle\int x^2(x^3 - 1)^4\,dx$.

The derivative of $x^3 - 1$ is $3x^2$, so write the integral as $\tfrac13\displaystyle\int 3x^2(x^3 - 1)^4\,dx$.

$\tfrac13 \times \dfrac{(x^3 - 1)^5}{5} + C = \dfrac{(x^3 - 1)^5}{15} + C$

Example 3●●

Evaluate $\displaystyle\int_0^4 \frac{x}{\sqrt{x^2 + 9}}\,dx$.

$\dfrac{x}{\sqrt{x^2 + 9}} = \tfrac12 \times 2x(x^2 + 9)^{-\frac12}$.

Primitive: $\tfrac12 \times \dfrac{(x^2 + 9)^{\frac12}}{\frac12} = \sqrt{x^2 + 9}$.

$\Big[\sqrt{x^2 + 9}\Big]_0^4 = 5 - 3 = 2$

Flashcards

Reverse chain rule

$\displaystyle\int f'(x)\,[f(x)]^n\,dx = \dfrac{[f(x)]^{n+1}}{n + 1} + C$

What to look for

The derivative of the inside of the bracket, multiplying the bracket.

A missing constant

Multiply inside and divide outside by it.

Practice

$\displaystyle\int 2x(x^2 + 1)^3\,dx = $
Evaluate $\displaystyle\int_0^1 2x(x^2 + 1)^3\,dx$.
$\displaystyle\int x^2(x^3 - 1)^4\,dx = $
$\displaystyle\int \frac{x}{\sqrt{x^2 + 9}}\,dx = $
Evaluate $\displaystyle\int_0^4 \frac{x}{\sqrt{x^2 + 9}}\,dx$.
Evaluate $\displaystyle\int_0^1 x(x^2 + 1)^2\,dx$.
Evaluate $\displaystyle\int_0^1 3x^2(x^3 + 1)^2\,dx$.
$\displaystyle\int (2x + 1)(x^2 + x)^5\,dx = $
Before using the reverse chain rule, check that: