Partial Integration and the Substitution Rule

I only want to write this down to have a place to look it up:

If we want to integrate ${\displaystyle {\int }_a^b{f}^{\prime }\left(x\right)g\left(x\right)}$ , we can use: $$\begin{array}{rl}(f\cdot g{)}^{\prime }\left(x\right)& =f^{\prime }\left(x\right)g\left(x\right)+f\left(x\right)g^{\prime }\left(x\right)\\ \iff f\left(x\right)g\left(x\right){|}_{a=x}^b& ={\int }_a^b{f}^{\prime }\left(x\right)g\left(x\right)+f\left(x\right)g^{\prime }\left(x\right)dx\\ \iff {\int }_a^b{f}^{\prime }\left(x\right)g\left(x\right)dx& =f\left(x\right)g\left(x\right){|}_{a=x}^b-{\int }_a^{b}f\left(x\right)g^{\prime }\left(x\right)dx\end{array}$$

And if we want to substitute a function in an integral:

$$\begin{gathered} {\int }_{g\left(a\right)}^{g\left(b\right)}f\left(x\right)dx=F\left(g\left(b\right)\right)-F\left(g\left(a\right)\right)=(F\circ g)\left(b\right)-(F\circ g)\left(a\right)= \\ ={\int }_a^b{(F\circ g)}^{\prime }\left(x\right)dx={\int }_a^{b}f\left(g\left(x\right)\right)g^{\prime }\left(x\right)dx \end{gathered}$$

(with $F\left(x\right)={\int }^{x}f\left(x\right)dx$) or using a trick:

$$\begin{gathered} {\int }_a^{b}f\left(g\left(x\right)\right)dx=F\left(b\right)-F\left(a\right)=(F\circ g\circ g^{-1})\left(b\right)-(F\circ g\circ g^{-1})\left(a\right)= \\ =(F\circ g^{-1})\left(g\left(b\right)\right)-(F\circ g^{-1})\left(g\left(a\right)\right)={\int }_{g\left(a\right)}^{g\left(b\right)}{(F\circ g^{-1})}^{\prime }\left(x\right)dx= \\ ={\int }_{g\left(a\right)}^{g\left(b\right)}{F}^{\prime }\left(g^{-1}\left(x\right)\right){g^{-1}}^{\prime }\left(x\right)dx={\int }_{g\left(a\right)}^{g\left(b\right)}f\left(g\left(g^{-1}\left(x\right)\right)\right){g^{-1}}^{\prime }\left(x\right)dx= \\ ={\int }_{g\left(a\right)}^{g\left(b\right)}f\left(x\right)\frac{1}{g^{\prime }\left(g^{-1}\left(x\right)\right)}dx \end{gathered}$$

(with $F\left(x\right)={\int }^{x}f\left(g\left(x\right)\right)dx$)