Integralen av $$$\frac{3 x}{\sqrt{1 - 9 x^{2}}}$$$

Bestäm $$$\int \frac{3 x}{\sqrt{1 - 9 x^{2}}}\, dx$$$.

Låt $$$u=1 - 9 x^{2}$$$ vara.

Då $$$du=\left(1 - 9 x^{2}\right)^{\prime }dx = - 18 x dx$$$ (stegen kan ses »), och vi har att $$$x dx = - \frac{du}{18}$$$.

Alltså,

$${\color{red}{\int{\frac{3 x}{\sqrt{1 - 9 x^{2}}} d x}}} = {\color{red}{\int{\left(- \frac{1}{6 \sqrt{u}}\right)d u}}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=- \frac{1}{6}$$$ och $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$:

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

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

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

Kom ihåg att $$$u=1 - 9 x^{2}$$$:

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

Alltså,

$$\int{\frac{3 x}{\sqrt{1 - 9 x^{2}}} d x} = - \frac{\sqrt{1 - 9 x^{2}}}{3}$$

Lägg till integrationskonstanten:

$$\int{\frac{3 x}{\sqrt{1 - 9 x^{2}}} d x} = - \frac{\sqrt{1 - 9 x^{2}}}{3}+C$$

$$$\int \frac{3 x}{\sqrt{1 - 9 x^{2}}}\, dx = - \frac{\sqrt{1 - 9 x^{2}}}{3} + C$$$A