The integral of a composite function times its inner derivative

By the chain rule, $\displaystyle {\left(F(u(x))\right)'}={f(u(x))\cdot u'(x)}$. So when the integral holds a composite function times the derivative of its inner function, the antiderivative is a composite function:

The common case – a power of a function times its derivative:

In the examples below, find in each integral the inner function $\displaystyle u$, and see that its derivative is already a factor:

• $\displaystyle \int2x\left(x^2+1\right)^3\,dx$: $\displaystyle {u}={x^2+1}$, $\displaystyle {u'}={2x}$. The result: $\displaystyle \frac{\left(x^2+1\right)^4}{4}+C$.
• $\displaystyle \int\frac{2x+1}{\left(x^2+x\right)^2}\,dx$: $\displaystyle {u}={x^2+x}$, $\displaystyle {u'}={2x+1}$, and $\displaystyle \frac{1}{u^2}$ gives $\displaystyle -\frac{1}{u}$. The result: $\displaystyle -\frac{1}{x^2+x}+C$.
• The check for the first: $\displaystyle {\left(\frac{\left(x^2+1\right)^4}{4}\right)'}={\frac{4\left(x^2+1\right)^3\cdot2x}{4}}={2x\left(x^2+1\right)^3}$.

When the inner derivative appears up to a number, make up the number: multiply by it inside the integral and divide by it outside. The example below finds $\displaystyle \int x^2\left(x^3+2\right)^4\,dx$:

• $\displaystyle {u}={x^3+2}$, $\displaystyle {u'}={3x^2}$, but the integral holds $\displaystyle x^2$ – a 3 is missing.
• $\displaystyle {\int x^2\left(x^3+2\right)^4\,dx}={\frac{1}{3}\int3x^2\left(x^3+2\right)^4\,dx}={\frac{1}{3}\cdot\frac{\left(x^3+2\right)^5}{5}+C}={\frac{\left(x^3+2\right)^5}{15}+C}$.
• In the same way, $\displaystyle {\int\frac{x}{\sqrt{x^2+1}}\,dx}={\frac{1}{2}\int\frac{2x}{\sqrt{x^2+1}}\,dx}={\frac{1}{2}\cdot2\sqrt{x^2+1}+C}={\sqrt{x^2+1}+C}$.
• A number missing – make it up. An $\displaystyle x$ missing – the rule does not work: expand $\displaystyle \int\left(x^3+2\right)^4\,dx$ instead.

The example below calculates $\displaystyle \int_0^1x\sqrt{1-x^2}\,dx$:

• $\displaystyle {u}={1-x^2}$, $\displaystyle {u'}={-2x}$: $\displaystyle {\int x\sqrt{1-x^2}\,dx}={-\frac{1}{2}\cdot\frac{2}{3}\left(1-x^2\right)\sqrt{1-x^2}+C}={-\frac{1}{3}\left(1-x^2\right)\sqrt{1-x^2}+C}$.
• $\displaystyle {\left[-\frac{1}{3}\left(1-x^2\right)\sqrt{1-x^2}\right]_0^1}={0-\left(-\frac{1}{3}\right)}={\frac{1}{3}}$.

Find the integrals.

Calculate the integrals.

For each integral, write whether the inner derivative appears in it, appears up to a number, or is missing:
$\displaystyle \int x\left(x^2-5\right)^3\,dx$, $\displaystyle \int\left(x^2-5\right)^3\,dx$, $\displaystyle \int\frac{x^2}{\sqrt{x^3+1}}\,dx$, $\displaystyle \int\frac{x}{\sqrt{x^3+1}}\,dx$.