Question

Evaluate:

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

Answer 1

$I=\int \frac{(\frac{1}{x^2}+1)dx}{(\frac{1}{x}-x)\sqrt{{(x-\frac{1}{x})}^2+2}}$

Putting $m=x-\frac{1}{x}$

$\Rightarrow dm=(1+\frac{1}{x^2})dx$

$\Rightarrow I=-\int \frac{dm}{m\sqrt{m^2+2}}$

Let $t=\sqrt{m^2+2}$

$\Rightarrow dt=\frac{mdm}{\sqrt{m^2+2}}\Rightarrow \frac{dt}{m}=\frac{dm}{\sqrt{m^2+2}}$

$I=-\int \frac{dt}{t^2-2}$

$=-\int \frac{dt}{(t+\sqrt{2})(t-\sqrt{2})}=\frac{-1}{2^{3/2}}\int [\frac{1}{t-\sqrt{2}}-\frac{1}{t+\sqrt{2}}]dt$

$=\frac{-1}{2^{3/2}}[\ln |t-\sqrt{2}|-\ln |t+\sqrt{2}|]+c$

$=\frac{1}{2^{3/2}}\ln |t+\sqrt{2}|-\frac{1}{2^{3/2}}\ln |t-\sqrt{2}|+c$

$=\frac{1}{2^{3/2}}\ln |\sqrt{m^2+2}+\sqrt{2}|-\frac{1}{2^{3/2}}\ln |\sqrt{m^2+2}-\sqrt{2}|+c$

$m^2+2={(x-\frac{1}{x})}^2+2$

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

$I=\frac{1}{2^{3/2}}\ln |\sqrt{x^2+\frac{1}{x^2}}+\sqrt{2}|-\frac{1}{2^{3/2}}\ln |\sqrt{x^2+\frac{1}{x^2}}-\sqrt{2}|+c$