finding value of $ \int\frac{1}{x^3\sqrt{(1+x)^3}}dx$

Question

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

Answer 1

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

Answer 2

A big partial fraction decomposition!

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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.

Answer 3

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 $$I=2\times\int\left(-\frac{1}{u^2}-\frac{15}{16(u+1)}-\frac{7}{16(u+1)^2}-\frac{1}{8(u+1)^3}+\frac{15}{16(u-1)}-\frac{7}{16(u-1)^2}+\frac{1}{8(u-1)^3}\right) du$$