<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] $$
\left.\begin{aligned}
\int \frac{\cos ^6 \theta}{\sin ^5 \theta}&=-\frac{1}{4} \int \cos ^5 \theta d\left(\frac{1}{\sin ^4 \theta}\right) \\
& =-\frac{\cos ^5 \theta}{4 \sin ^4 \theta}-\frac{5}{4} \int \frac{\cos ^4 \theta \sin ^2 d \theta}{\sin ^4 \theta} \\
& =-\frac{\cos ^5 \theta}{4 \sin ^4 \theta}-\frac{5}{4} \int \frac{\cos ^4 \theta}{\sin ^3 \theta} d \theta \\
& =-\frac{\cos ^5 \theta}{4 \sin ^4 \theta}+\frac{5}{8} \int \cos ^3 \theta d\left(\frac{1}{\sin ^2 \theta}\right) \\
& =-\frac{\cos ^5 \theta}{4 \sin ^4 \theta}+\frac{5 \cos ^3 \theta}{8 \sin ^2 \theta}+\frac{15}{8} \int \frac{\cos ^2 \theta}{\sin \theta} d \theta \\
& =-\frac{\cos ^5 \theta}{4 \sin ^4 \theta}+\frac{5 \cos ^3 \theta}{8 \sin ^2 \theta}+\frac{15}{8} \int \frac{1-\sin \theta}{\sin \theta} d \theta \\
& =-\frac{\cos ^5 \theta}{4 \sin ^4 \theta}+\frac{5 \cos ^3 \theta}{8 \sin ^2 \theta}-\frac{15}{8}[\ln |\csc \theta+\cot \theta|-\cos \theta ]+C\\
& = \frac{1}{4 x^2 \sqrt{1+x}}+\frac{5}{8 x} \sqrt{1+x}-\frac{15}{8} \ln \left|\frac{\sqrt{1+x}+1}{\sqrt{x}}\right|-\frac{1}{\sqrt{1+x}}+C
\end{aligned}\right]
$$
$$
\begin{aligned}
I=&-\frac{1}{2 x^2 \sqrt{1+x}}+\frac{5}{4 x \sqrt{1+x}}-\frac{15}{4} \ln \left|\frac{\sqrt{1+x}+1}{\sqrt{x}}\right| -\frac{1}{2 \sqrt{1+x}}+C
\end{aligned}
$$
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