Integral of $\arcsin(1/x)$

Exercise:

Integrate, $\int{\arcsin \left( \frac{1}{x} \right)dx}$

Solution: Let $u=\arcsin \left( \frac{1}{x} \right)\,\,\,\And \,\,\,dv=\int{dx}$

$\Rightarrow du=\frac{1}{\sqrt{1-\frac{1}{{{x}^{2}}}}}\times \left( \frac{-1}{{{x}^{2}}} \right)dx\,\,\,\,\,\And \,\,\,\,v=x$

So $\int{\arcsin \left( \frac{1}{x} \right)dx}=x\arcsin \left( \frac{1}{x} \right)+\int{x\times \frac{dx}{{{x}^{2}}\sqrt{1-\frac{1}{{{x}^{2}}}}}}$

But $\int{x\times \frac{dx}{{{x}^{2}}\sqrt{1-\frac{1}{{{x}^{2}}}}}=\int{\frac{dx}{x\sqrt{1-\frac{1}{{{x}^{2}}}}}}}=\int{\frac{dx}{x\sqrt{\frac{{{x}^{2}}-1}{{{x}^{2}}}}}}=\int{\frac{dx}{x\frac{\sqrt{{{x}^{2}}-1}}{x}}=\int{\frac{dx}{\sqrt{{{x}^{2}}-1}}}}$

Let ${{u}^{2}}={{x}^{2}}-1\Rightarrow 2udu=2xdx\Rightarrow dx=\frac{u}{x}du\,\,\And \,\,\,x=\sqrt{{{u}^{2}}+1}$

So $\int{\frac{dx}{\sqrt{{{x}^{2}}-1}}=\int{\frac{1}{u}\times \frac{udu}{\sqrt{{{u}^{2}}+1}}=\int{\frac{du}{\sqrt{{{u}^{2}}+1}}}}}$

Let $u=\tan \theta \Rightarrow du={{\sec }^{2}}\theta \,d\theta $ and ${{\sec }^{2}}\theta ={{\tan }^{2}}\theta +1={{u}^{2}}+1\Rightarrow \sec \theta =\sqrt{{{u}^{2}}+1}$

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

$=\ln \left| \sec \theta +\tan \theta  \right|+c=\ln \left( u+\sqrt{{{u}^{2}}+1} \right)+c=\ln \left( \sqrt{{{x}^{2}}-1}+x \right)+c$

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