How do I calculate the following integral $$\int\frac{dx}{x^{3}(x^{2}+1)\sqrt{x^{2}+1}}$$ I found this result in a book $$\int\frac{dx}{x^{3}(x^{2}+1)\sqrt{x^{2}+1}}=\frac{-1}{2x^{2}\sqrt{x^{2}+1}}-\frac{3}{2\sqrt{x^{2}+1}}+\frac{3}{2}\ln(\frac{1+\sqrt{x^{2}+1}}{x})+C$$ but I couldn't find a way to solve it
Despite trying, please help me to find a way so that I can understand how to solve it
Thanks in advance
On
$$ \begin{aligned} \int\frac{1}{x^{3}(x^{2}+1)\sqrt{x^{2}+1}}dx &\stackrel{t=\frac{1}{x}}{=}-\int \frac{t^4}{\left(1+t^2\right)^{\frac{3}{2}}} d x \\ &=\int t^3 d\left(\frac{1}{\sqrt{1+t^2}}\right) \\ &=\frac{t^3}{\sqrt{1+t^2}}-3 \int \frac{t^2}{\sqrt{1+t^2}} d t \quad \textrm{( via IBP)} \end{aligned} $$
Letting $x=\sinh \theta$ yields $$ \begin{aligned} \int \frac{t^2}{\sqrt{1+t^2}} d t&=\int \sinh ^2 \theta d \theta=\int \frac{\cosh 2 \theta-1}{2} d \theta =\frac{\sinh \theta}{4}-\frac{\theta}{2}= \frac{1}{2}\left(t \sqrt{1+t^2}-\sinh ^{-1} t\right) \end{aligned} $$
Putting all back yields $$ \int\frac{1}{x^{3}(x^{2}+1)\sqrt{x^{2}+1}}dx = \frac{2}{3}\left[-\frac{x^2+1}{x^2 \sqrt{x^2+1}}+\sinh ^{-1}\left(\frac{1}{x}\right)\right]+C $$
On
$$I=\int\frac{dx}{x^{3}(x^{2}+1)\sqrt{x^{2}+1}}$$ Let $x=\sinh(t)$ $$I=\int \text{csch}^3(t)\, \text{sech}^2(t)\,dt$$ Let $t=2 \tanh^{-1}(u)$ $$I=\int\frac{\left(u^2-1\right)^4}{4 u^3 \left(u^2+1\right)^2}\,du$$ Using partial fraction decomposition $$\frac{\left(u^2-1\right)^4}{4 u^3 \left(u^2+1\right)^2}= \frac{1}{4 u^3}+\frac{4 u}{\left(u^2+1\right)^2}+\frac{u}{4}-\frac{3}{2 u}$$ and all antiderivatives are simple.
Shorter would be to use directly $$x=\sinh \left(2 \tanh ^{-1}(u)\right)=\frac {2u}{1-u^2}$$ for the same result.
Integrate by parts as follows
\begin{align} &\int\frac{1}{x^{3}(x^{2}+1)\sqrt{x^{2}+1}}\ dx\\ =& -\int \frac{1}{2(x^2+1) \sqrt{x^{2}+1}}\ d\left(1+\frac1{x^2}\right)\\ \overset{ibp}=& \ -\frac{1}{2x^2\sqrt{x^{2}+1}} -\frac32 \int \frac1x\ d\bigg( \frac{x}{\sqrt{x^{2}+1}}\bigg)\\ \overset{ibp}= & \ -\frac{1}{2x^2\sqrt{x^{2}+1}}-\frac32 \bigg( \frac{1}{\sqrt{x^{2}+1}}-\int \frac1{x \sqrt{x^{2}+1}}dx\bigg)\\ = & \ -\frac{1}{2x^2\sqrt{x^{2}+1}}- \frac{3}{2\sqrt{x^{2}+1}}+\frac32 \tanh^{-1}\sqrt{x^2+1} \end{align}