Find the indefinite integral $\int {dx \over {(1+x^2) \sqrt{1-x^2}}} $ Find the indefinite integral
$$\int {dx \over {(1+x^2) \sqrt{1-x^2}}} $$
Is there a smart substitution or algebric trick that I'm missing? Because integration by parts hasn't helped..

Let $$I = \int\frac{1}{(1+x^2)\sqrt{1-x^2}}dx\;,$$ Now Put $\displaystyle x= \frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt$
So we get $$I = \int\frac{t^3}{(1+t^2)\sqrt{t^2-1}}\cdot -\frac{1}{t^2}dt = -\int\frac{t}{(1+t^2)\sqrt{t^2-1}}dt$$
Now Put $(t^2-1)=u^2\;,$ Then $2tdt = 2udu\Rightarrow tdt = udu$
So we get $$I = -\int\frac{u}{u^2+2}\cdot \frac{1}{u}du = -\int\frac{1}{u^2+2}du$$
So we get $$I = -\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{u}{\sqrt{2}}\right)+\mathcal{C}=-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{t^2-1}}{\sqrt{2}}\right)+\mathcal{C}$$
So we get $$I = \int\frac{1}{(1+x^2)\sqrt{1-x^2}}dx=-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{1-x^2}}{\sqrt{2}x}\right)+\mathcal{C}$$
