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If $F'(x)=f(x)$, then $(F(g(x)))'=f(g(x))g'(x)$. Thus, the integral of a function of the type

$$\int f(g(x))g'(x)\,\mathrm{d}x$$

is always solved as

$$\int f(g(x))g'(x)\,\mathrm{d}x=F(g(x))+k$$

It is very useful to denote this trick with a change of variable: let

$$t=g(x)$$

If we differentiate both sides of the expression and keep track of variable with respect to which we are differentiating with $\mathrm{d}\cdot$, we get

$$\mathrm{d}t = g'(x)\,\mathrm{d}x$$

and the expression above is more easily handed

$$\int f(g(x))g'(x)\,\mathrm{d}x\,\overset{\substack{g(x)=t\\g'(x)\,\mathrm{d}x=\mathrm{d}t\\ \phantom{\downarrow}}}{=}\,\int f(t)\,\mathrm{d}t=F(t)+k\,\overset{\substack{t=g(x)\\ \phantom{\downarrow}}}{=}\,F(g(x))+k$$

This also leads to a list of "semi-immediate integrals": if $t$ is some function of $x$,

$$ \int t^n\cdot t' \,\mathrm{d}x=\dfrac{t^{n+1}}{n+1}+k\qquad \text{if $n\neq -1$} $$$$ \int \dfrac{t'}{t}\,\mathrm{d}x=\ln\vert t\vert+k $$
$$ \int e^t\cdot t' \,\mathrm{d}x= e^t+k $$$$ \int a^t\cdot t' \,\mathrm{d}x= \dfrac{a^t}{\ln a}+k $$
$$ \int \sin t\cdot t' \,\mathrm{d}x= -\cos t+k $$$$ \int \cos t\cdot t' \,\mathrm{d}x= \sin t+k $$
$$ \int \dfrac{t'}{1+t^2} \,\mathrm{d}x = \tan^{-1} t+k $$$$ \int \left(1+\tan^2 t\right)\cdot t' \,\mathrm{d}x= \int \dfrac{t'}{\cos^2 t} \,\mathrm{d}x = \tan t+k $$
$$ \dfrac{t'}{\sqrt{1-t^2}} \,\mathrm{d}x = \sin^{-1} t+k $$$$ \dfrac{t'}{\sqrt{1+t^2}} \,\mathrm{d}x = \sinh^{-1} t+k $$