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

Question

The standard operating procedure for such questions is substituting $x$ with $\frac 1t$

So $$\int \frac{-t}{(t^2+1)\sqrt {t^2-1}} \text dt$$

Which gives more or less the same expression. How do I solve this question?

Answer 1

You are almost there !

we have $$\int \frac{-t}{(t^2+1)(\sqrt {t^2-1})} dt$$ let $t^2=u$ :then we have $$\int \frac{-du}{2(u+1)\sqrt{(u-1)}}$$ Now take $\sqrt{u-1}=y$ we have $$\int \frac{-dy}{y^2+2}$$ which should be easy

Answer 2

Let $x=\sin(t)$, so $dx=\cos(t)dt$. Then $\sqrt{1-x^{2}}=\sqrt{1-\sin^{2}(t)}=\cos(t)$ and $t=\arcsin(x)$, so we can see that $$\int \frac{1}{\sqrt{1-x^{2}}(x^{2}+1)}dx=\int \frac{1}{\sin^{2}(t)+1}dt=\int \frac{1}{\sin^{2}(t)+1}\left(\frac{\csc^{2}(t)}{\csc^{2}(t)}\right)dt$$ now, let $s=\cot(t)$ then $ds=-\csc^{2}(t)$. Therefore, $$\int \frac{1}{\sin^{2}(t)+1}\left(\frac{\csc^{2}(t)}{\csc^{2}(t)}\right)dt=\int-\frac{1}{s^{2}+2}ds=-\frac{1}{\sqrt{2}}\arctan \left(\frac{s}{\sqrt{2}}\right)+c$$ So, we have $$\boxed{\int \frac{1}{\sqrt{1-x^{2}}(x^{2}+1)}dx =\frac{1}{\sqrt{2}}\arctan \left(\frac{\sqrt{2}x}{\sqrt{1-x^{2}}} \right)+c}$$