integral exercise asked in the mathematics teachers group

Exercise:

Integrate, $\int{\frac{x\arctan x}{\sqrt{{{\left( 1+{{x}^{2}} \right)}^{3}}}}dx}$

Solution: Let $x=\tan \theta \Leftrightarrow dx={{\sec }^{2}}\theta d\theta $

So $\int{\frac{x\arctan x}{\sqrt{{{\left( 1+{{x}^{2}} \right)}^{3}}}}dx}=\int{\frac{\tan \theta \arctan \left( \tan \theta  \right)}{\sqrt{{{\left( 1+{{\tan }^{2}}\theta  \right)}^{3}}}}}{{\sec }^{2}}\theta d\theta $

$=\int{\frac{\theta \tan \theta }{\sqrt{{{\left( {{\sec }^{2}}\theta  \right)}^{3}}}}{{\sec }^{2}}\theta d\theta }=\int{\frac{\theta \tan \theta }{{{\sec }^{3}}\theta }{{\sec }^{2}}\theta d\theta }=\int{\frac{\theta \tan \theta }{\sec \theta }d\theta }$

But $\frac{\tan \theta }{\sec \theta }=\frac{\frac{\sin \theta }{\cos \theta }}{\frac{1}{\cos \theta }}=\sin \theta $

So $\int{\frac{\theta \tan \theta }{\sec \theta }d\theta }=\int{\theta \sin \theta d\theta }$

Let $u=\theta \,\,\And \,\,dv=\sin \theta d\theta \Leftrightarrow du=d\theta \,\,\And \,\,v=-\cos \theta $

Hence $\int{\theta \sin \theta \,d\theta }=-\theta \cos \theta +\int{\cos \theta d\theta }=-\theta \cos \theta +\sin \theta +c$

Now by Backward substitution we get :

We have $x=\tan \theta \Rightarrow \theta =\arctan x$ and $\cos \theta =\frac{1}{\sqrt{1+{{x}^{2}}}}\,\,\And \,\,\sin \theta =\frac{x}{\sqrt{1+{{x}^{2}}}}$

Thus $\int{\frac{x\arctan x}{\sqrt{{{\left( 1+{{x}^{2}} \right)}^{3}}}}dx}=\frac{-\arctan x}{\sqrt{1+{{x}^{2}}}}+\frac{x}{\sqrt{1+{{x}^{2}}}}+c=\frac{x-\arctan x}{\sqrt{1+{{x}^{2}}}}+c$