Question

$\int \frac{x^2}{\sqrt{x-8}}dx.$

Answer 1

$I=\int \frac{x^2}{\sqrt{x-8}}dx=\int (\frac{x^2-64}{\sqrt{x-8}}+\frac{64}{\sqrt{x-8}})dx$

$=\int \left(\sqrt{x-8}\right)(x+8)dx+64\times 2\sqrt{x-8}$

put x = $8se{c}^2\theta \Rightarrow tan\theta =\frac{\sqrt{x-8}}{\sqrt{8}}$

$\Rightarrow dx=16se{c}^2\theta tan\theta d\theta$

$=\int \sqrt{8}tan\theta \sqrt{8}(se{c}^2\theta +1)\times 16se{c}^2\theta tan\theta d\theta +128\sqrt{x-8}$

put $tan\theta =t$

$sec\theta d\theta =dt$

$\int 128\sqrt{8}{t}^2(2+t^2)+dt+128\sqrt{x-8}$

$\int 128\sqrt{8}(t^4+2t^2)dt+128\sqrt{x-8}$

$128\sqrt{8}(\frac{t^5}{5}+\frac{2t^3}{3})+128\sqrt{x-8}$

$=128\sqrt{8}(\frac{tan\theta 5}{5}+\frac{2ta{n}^3\theta }{3})+128\sqrt{x-8}$

$I=128\sqrt{8}\left[\right(\frac{\sqrt{x-8}}{\sqrt{8}}{)}^5\times \frac{1}{5}+\frac{2}{3}\left(\frac{\sqrt{x-8}}{\sqrt{8}}{)}^3\right]+128\sqrt{x-8}$