Two integral exercise for rational and irrational

Exercise:

Integrate $\int{\frac{dx}{{{x}^{6}}+x}}$

Solution: we have $\frac{1}{{{x}^{6}}+x}=\frac{1}{{{x}^{6}}\left( 1+{{x}^{-5}} \right)}=\frac{{{x}^{-6}}}{1+{{x}^{-5}}}$ 

Let $u=1+{{x}^{-5}}\Rightarrow du=-5{{x}^{-6}}dx$

So $\int{\frac{dx}{{{x}^{6}}+x}}=\int{\frac{{{x}^{-6}}}{1+{{x}^{-5}}}dx}=-\frac{1}{5}\int{\frac{du}{u}=-\frac{1}{5}\ln \left| u \right|+c}=-\frac{1}{5}\ln \left| 1+\frac{1}{{{x}^{5}}} \right|+c$

Exercise:

Integrate, $\int{\frac{dx}{{{x}^{2}}\sqrt{{{x}^{2}}-1}}}$

Solution: Let $x=\sec \theta \Leftrightarrow dx=\sec \theta \tan \theta d\theta $

So $\int{\frac{dx}{{{x}^{2}}\sqrt{{{x}^{2}}-1}}=\int{\frac{\sec \theta \tan \theta d\theta }{{{\sec }^{2}}\theta \sqrt{{{\sec }^{2}}\theta -1}}=\int{\frac{\tan \theta }{\sec \theta \tan \theta }d\theta =\int{\frac{1}{\sec \theta }d\theta }=\int{\cos \theta }}\,d\theta }}$

$=\sin \theta +c$ but $x=\frac{1}{\cos \theta }\Rightarrow \cos \theta =\frac{1}{x}$ by Pythagoras identity $\sin \theta =\sqrt{1-\frac{1}{{{x}^{2}}}}$

$=\sqrt{1-\frac{1}{{{x}^{2}}}}+c=\sqrt{\frac{{{x}^{2}}-1}{{{x}^{2}}}}+c=\frac{\sqrt{{{x}^{2}}-1}}{x}+c$