Question Number 51456 by Tawa1 last updated on 27/Dec/18
$$\int\:\:\:\frac{\mathrm{x}^{\mathrm{8}} }{\mathrm{x}^{\mathrm{6}} \:+\:\mathrm{64}}\:\mathrm{dx} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Dec/18
![x^3 =2^3 tanθ x=2(tanθ)^(1/3) dx=(2/3)×(tanθ)^((−2)/3) ×sec^2 θ ∫((2^8 ×(tanθ)^(8/3) ×(2/3)×(tanθ)^((−2)/3) sec^2 θ)/(2^6 tan^2 θ+2^6 ))dθ =(2^3 /3)∫(tanθ)^((8/3)−(2/3)) dθ =(8/3)∫tan^2 θdθ =(8/3)[∫(sec^2 θ−1)dθ =(8/3)[tanθ−θ]+c =(8/3)[(x^3 /8)−tan^(−1) ((x^3 /8))]+c](Q51468.png)
$${x}^{\mathrm{3}} =\mathrm{2}^{\mathrm{3}} {tan}\theta \\ $$$${x}=\mathrm{2}\left({tan}\theta\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${dx}=\frac{\mathrm{2}}{\mathrm{3}}×\left({tan}\theta\right)^{\frac{−\mathrm{2}}{\mathrm{3}}} ×{sec}^{\mathrm{2}} \theta \\ $$$$\int\frac{\mathrm{2}^{\mathrm{8}} ×\left({tan}\theta\right)^{\frac{\mathrm{8}}{\mathrm{3}}} ×\frac{\mathrm{2}}{\mathrm{3}}×\left({tan}\theta\right)^{\frac{−\mathrm{2}}{\mathrm{3}}} {sec}^{\mathrm{2}} \theta}{\mathrm{2}^{\mathrm{6}} {tan}^{\mathrm{2}} \theta+\mathrm{2}^{\mathrm{6}} }{d}\theta \\ $$$$=\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}}\int\left({tan}\theta\right)^{\frac{\mathrm{8}}{\mathrm{3}}−\frac{\mathrm{2}}{\mathrm{3}}} {d}\theta \\ $$$$=\frac{\mathrm{8}}{\mathrm{3}}\int{tan}^{\mathrm{2}} \theta{d}\theta \\ $$$$=\frac{\mathrm{8}}{\mathrm{3}}\left[\int\left({sec}^{\mathrm{2}} \theta−\mathrm{1}\right){d}\theta\right. \\ $$$$=\frac{\mathrm{8}}{\mathrm{3}}\left[{tan}\theta−\theta\right]+{c} \\ $$$$=\frac{\mathrm{8}}{\mathrm{3}}\left[\frac{{x}^{\mathrm{3}} }{\mathrm{8}}−{tan}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{3}} }{\mathrm{8}}\right)\right]+{c} \\ $$
Commented by Tawa1 last updated on 27/Dec/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$