Integral exercise Level 1

Exercise:

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

Solution: First take $u=\ln x\Leftrightarrow du=\frac{1}{x}dx\,\,$

So $\int{\frac{dx}{x\sqrt{{{\left( 1+{{\ln }^{2}}x \right)}^{3}}}}=\int{\frac{dx}{x}\times \frac{1}{\sqrt{{{\left( 1+{{\ln }^{2}}x \right)}^{3}}}}=\int{\frac{du}{\sqrt{{{\left( 1+{{u}^{2}} \right)}^{3}}}}}}}$

Now Take $u=\tan \theta \Leftrightarrow du={{\sec }^{2}}\theta \,d\theta $

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

But $\sec \theta =\frac{1}{\cos \theta }\Leftrightarrow \frac{1}{\sec \theta }=\cos \theta $

Thus $\int{\frac{d\theta }{\sec \theta }=\int{\cos \theta \,d\theta }=\sin \theta +c}$

But $u=\tan \theta \Leftrightarrow \theta =\arctan \left( u \right)=\arctan \left( \ln x \right)$

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