<s>[INST] finding value of $ \int\frac{1}{x^3\sqrt{(1+x)^3}}dx$ finding value of $\displaystyle \int\frac{1}{x^3\sqrt{(1+x)^3}}dx$
Substituting $x=\tan^2 \theta\;, dx = 2\tan \theta \sec^2 \theta$
integral is $=\displaystyle \int \frac{2\tan \theta \sec^2 \theta }{\tan^6 \theta \cdot \sec^3 \theta}d\theta= 2\int\frac{\cos^6 \theta}{\sin^5 \theta}d\theta $
wan,t be able to proceed after, could some help me
 [/INST] A big partial fraction decomposition!

We set $u =\cos x$ to get $$I = -\int \frac{\cos ^6 x}{(\cos ^2x-1)^3} \sin x dx = \int \frac{u^6}{(u^2-1)^3} du =\int (\frac{3u^4-3u^2+1}{(u^2-1)^3} + 1) du = I_1 + I_2$$
For $I_1$, perform the partial fraction decomposition $$I_1 = \int \frac{3u^4-3u^2+1}{(u-1)^3(u+1)^3} du =-\int \frac{15}{16(u+1)} du +\int \frac{9}{16(u+1)^2} du -\int \frac{1}{8(u+1)^3} du +\int \frac{15}{16(u-1)} du + \int \frac{9}{16(u-1)^2} du +\int \frac{1}{8(u-1)^3} du $$ The integration of $I_2$ is however very easy. Hope you can take it from here. 
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