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 $x=\sin\theta\implies dx=\cos\theta \ d\theta$, $$\int \frac{dx}{(1+x^2)\sqrt{1-x^2}}$$ $$=\int \frac{\cos\theta\ d\theta}{(1+\sin^2\theta)\cos\theta}$$
$$=\int \frac{\sec^2\theta\ d\theta}{\sec^2\theta(1+\sin^2\theta)}$$
$$=\int \frac{\sec^2\theta\ d\theta}{\sec^2\theta+\tan^2\theta}$$
$$=\int \frac{\sec^2\theta\ d\theta}{1+2\tan^2\theta}$$
$$=\int \frac{ d(\tan\theta)}{1+(\sqrt2 \tan\theta)^2}$$
$$=\frac{1}{\sqrt 2}\tan^{-1}\left(\sqrt 2\tan\theta\right)+C$$
$$=\color{red}{\frac{1}{\sqrt 2}\tan^{-1}\left(\frac{x\sqrt 2}{\sqrt{1-x^2}}\right)+C}$$
