<s>[INST] 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..
 [/INST] Substitute $u = \frac{x\sqrt2}{\sqrt{1-x^2}}, \frac{dx}{du} =\frac{(1-x^2)^{3/2}}{\sqrt2} $
$\begin{align*}
\int {\frac1 {(1+x^2) \sqrt{1-x^2}}dx} &= \frac1{\sqrt2}\int  \frac{1-x^2}{1+x^2} du \\
&= \frac1{\sqrt2}\int \frac1 {\frac{1+x^2}{1-x^2}} du \\
&= \frac1{\sqrt2}\int \frac1 {1+\frac{2x^2}{1-x^2}} du \\
&= \frac1{\sqrt2}\int \frac1 {1+ (\frac{x\sqrt2}{\sqrt{1-x^2}})^2} du \\ 
&= \frac1{\sqrt2}\int \frac1 {1+ u^2} du \\
&= \frac1{\sqrt2} \tan^{-1} \left(\frac{x\sqrt2}{\sqrt{1-x^2}} \right) + C 
\end{align*}$
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