Integralen av $$$\frac{1}{\left(x - 1\right)^{3}}$$$

Bestäm $$$\int \frac{1}{\left(x - 1\right)^{3}}\, dx$$$.

Låt $$$u=x - 1$$$ vara.

Då $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (stegen kan ses »), och vi har att $$$dx = du$$$.

Integralen kan omskrivas som

$${\color{red}{\int{\frac{1}{\left(x - 1\right)^{3}} d x}}} = {\color{red}{\int{\frac{1}{u^{3}} d u}}}$$

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=-3$$$:

$${\color{red}{\int{\frac{1}{u^{3}} d u}}}={\color{red}{\int{u^{-3} d u}}}={\color{red}{\frac{u^{-3 + 1}}{-3 + 1}}}={\color{red}{\left(- \frac{u^{-2}}{2}\right)}}={\color{red}{\left(- \frac{1}{2 u^{2}}\right)}}$$

Kom ihåg att $$$u=x - 1$$$:

$$- \frac{{\color{red}{u}}^{-2}}{2} = - \frac{{\color{red}{\left(x - 1\right)}}^{-2}}{2}$$

Alltså,

$$\int{\frac{1}{\left(x - 1\right)^{3}} d x} = - \frac{1}{2 \left(x - 1\right)^{2}}$$

Lägg till integrationskonstanten:

$$\int{\frac{1}{\left(x - 1\right)^{3}} d x} = - \frac{1}{2 \left(x - 1\right)^{2}}+C$$

$$$\int \frac{1}{\left(x - 1\right)^{3}}\, dx = - \frac{1}{2 \left(x - 1\right)^{2}} + C$$$A