Integralen av $$$- \frac{6}{\left(2 x - 9\right)^{2}}$$$

Bestäm $$$\int \left(- \frac{6}{\left(2 x - 9\right)^{2}}\right)\, dx$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=-6$$$ och $$$f{\left(x \right)} = \frac{1}{\left(2 x - 9\right)^{2}}$$$:

$${\color{red}{\int{\left(- \frac{6}{\left(2 x - 9\right)^{2}}\right)d x}}} = {\color{red}{\left(- 6 \int{\frac{1}{\left(2 x - 9\right)^{2}} d x}\right)}}$$

Låt $$$u=2 x - 9$$$ vara.

Då $$$du=\left(2 x - 9\right)^{\prime }dx = 2 dx$$$ (stegen kan ses »), och vi har att $$$dx = \frac{du}{2}$$$.

Alltså,

$$- 6 {\color{red}{\int{\frac{1}{\left(2 x - 9\right)^{2}} d x}}} = - 6 {\color{red}{\int{\frac{1}{2 u^{2}} d u}}}$$

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

$$- 6 {\color{red}{\int{\frac{1}{2 u^{2}} d u}}} = - 6 {\color{red}{\left(\frac{\int{\frac{1}{u^{2}} d u}}{2}\right)}}$$

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

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

Kom ihåg att $$$u=2 x - 9$$$:

$$3 {\color{red}{u}}^{-1} = 3 {\color{red}{\left(2 x - 9\right)}}^{-1}$$

Alltså,

$$\int{\left(- \frac{6}{\left(2 x - 9\right)^{2}}\right)d x} = \frac{3}{2 x - 9}$$

Lägg till integrationskonstanten:

$$\int{\left(- \frac{6}{\left(2 x - 9\right)^{2}}\right)d x} = \frac{3}{2 x - 9}+C$$

$$$\int \left(- \frac{6}{\left(2 x - 9\right)^{2}}\right)\, dx = \frac{3}{2 x - 9} + C$$$A