Integralen av $$$\frac{1}{\left(g - 27\right)^{\frac{2}{3}}}$$$

Bestäm $$$\int \frac{1}{\left(g - 27\right)^{\frac{2}{3}}}\, dg$$$.

Låt $$$u=g - 27$$$ vara.

Då $$$du=\left(g - 27\right)^{\prime }dg = 1 dg$$$ (stegen kan ses »), och vi har att $$$dg = du$$$.

Integralen kan omskrivas som

$${\color{red}{\int{\frac{1}{\left(g - 27\right)^{\frac{2}{3}}} d g}}} = {\color{red}{\int{\frac{1}{u^{\frac{2}{3}}} d u}}}$$

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

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

Kom ihåg att $$$u=g - 27$$$:

$$3 \sqrt[3]{{\color{red}{u}}} = 3 \sqrt[3]{{\color{red}{\left(g - 27\right)}}}$$

Alltså,

$$\int{\frac{1}{\left(g - 27\right)^{\frac{2}{3}}} d g} = 3 \sqrt[3]{g - 27}$$

Lägg till integrationskonstanten:

$$\int{\frac{1}{\left(g - 27\right)^{\frac{2}{3}}} d g} = 3 \sqrt[3]{g - 27}+C$$

$$$\int \frac{1}{\left(g - 27\right)^{\frac{2}{3}}}\, dg = 3 \sqrt[3]{g - 27} + C$$$A