How would you integrate $\sqrt{1+\frac{1}{x^2}}$ I need to integrate $\sqrt{1+\frac{1}{x^2}}$
I've tried to let $u=1/x^2$ but end up with, $\int \frac{\sqrt{1+u^2}}{u^{3/2}}du$ . I attempted to then substitute $u=\tan\theta$ and lead me to $-\frac{1}{2}\int \frac{1}{(\sin\theta\cos\theta)^{3/2}}d\theta$.
Not sure how to progress from here. Is there any alternative way of doing this?

$$\int\sqrt{1+\frac{1}{x^2}}\,dx=\int \frac{\sqrt{1+x^2}}{x}\,dx=\int \frac{x\sqrt{1+x^2}}{x^2}\,dx$$ Let $1+x^2=u^2 $ so that$$\int \frac{x\sqrt{1+x^2}}{x^2}\,dx=\int \frac{u^2}{u^2-1}\,du$$ hence we have that $$\int \frac{u^2}{u^2-1}\,du=u+\frac{1}{2}\ln |\frac{u-1}{u+1}|+c$$
