Integralen av $$$\frac{1}{\left(3 x - 1\right)^{2}}$$$

Bestäm $$$\int \frac{1}{\left(3 x - 1\right)^{2}}\, dx$$$.

Låt $$$u=3 x - 1$$$ vara.

Då $$$du=\left(3 x - 1\right)^{\prime }dx = 3 dx$$$ (stegen kan ses »), och vi har att $$$dx = \frac{du}{3}$$$.

Alltså,

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{3}$$$ och $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$:

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

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=-2$$$:

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

Kom ihåg att $$$u=3 x - 1$$$:

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

Alltså,

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

Förenkla:

$$\int{\frac{1}{\left(3 x - 1\right)^{2}} d x} = - \frac{1}{9 x - 3}$$

Lägg till integrationskonstanten:

$$\int{\frac{1}{\left(3 x - 1\right)^{2}} d x} = - \frac{1}{9 x - 3}+C$$

$$$\int \frac{1}{\left(3 x - 1\right)^{2}}\, dx = - \frac{1}{9 x - 3} + C$$$A