Integralen av $$$\frac{x}{\left(64 - x^{2}\right)^{\frac{3}{2}}}$$$

Bestäm $$$\int \frac{x}{\left(64 - x^{2}\right)^{\frac{3}{2}}}\, dx$$$.

Låt $$$u=64 - x^{2}$$$ vara.

Då $$$du=\left(64 - x^{2}\right)^{\prime }dx = - 2 x dx$$$ (stegen kan ses »), och vi har att $$$x dx = - \frac{du}{2}$$$.

Integralen kan omskrivas som

$${\color{red}{\int{\frac{x}{\left(64 - x^{2}\right)^{\frac{3}{2}}} d x}}} = {\color{red}{\int{\left(- \frac{1}{2 u^{\frac{3}{2}}}\right)d u}}}$$

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

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

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=- \frac{3}{2}$$$:

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

Kom ihåg att $$$u=64 - x^{2}$$$:

$$\frac{1}{\sqrt{{\color{red}{u}}}} = \frac{1}{\sqrt{{\color{red}{\left(64 - x^{2}\right)}}}}$$

Alltså,

$$\int{\frac{x}{\left(64 - x^{2}\right)^{\frac{3}{2}}} d x} = \frac{1}{\sqrt{64 - x^{2}}}$$

Lägg till integrationskonstanten:

$$\int{\frac{x}{\left(64 - x^{2}\right)^{\frac{3}{2}}} d x} = \frac{1}{\sqrt{64 - x^{2}}}+C$$

$$$\int \frac{x}{\left(64 - x^{2}\right)^{\frac{3}{2}}}\, dx = \frac{1}{\sqrt{64 - x^{2}}} + C$$$A