<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] Here is another way...
First substitute $$u^2=1+x$$ so that $$I=2\int\frac{udu}{(u^2-1)^3u^3}$$
Then partial fraction decomposition gives $$I=2\times\int\left(-\frac{1}{u^2}-\frac{15}{16(u+1)}-\frac{7}{16(u+1)^2}-\frac{1}{8(u+1)^3}+\frac{15}{16(u-1)}-\frac{7}{16(u-1)^2}+\frac{1}{8(u-1)^3}\right) du$$
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