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

is always solved as

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

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

and the expression above is more easily handed

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

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