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Evaluate the integral:

$$\int_0^1 \frac{\sqrt{1-x^2}}{\left(1+x\right)^2}\, dx$$

\begin{align*}
\int_0^1 \frac{\sqrt{1-x^2}}{(1+x)^2}\, dx&=\int_0^1 \sqrt{\frac{1-x}{1+x}}\,\frac{1}{1+x}\, dx\\
&\mathop{=\!=\!=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,\sqrt{\frac{1-x}{1+x}}}\\
{dx\,=-\frac{4t}{(1+t^2)^2}\,dt} \\
\end{subarray}} \,-\int_1^0 t\,\frac{1}{1+\frac{1-t^2}{1+t^2}}\, \frac{4t}{(1+t^2)^2}\,dt\\
&=\int_0^1 \frac{2t^2}{1+t^2}\,dt\\
&=\int_0^1 \frac{2t^2+2-2}{1+t^2}\,dt\\
&=\int_0^1 2\,dt-2\int_0^1 \frac{1}{1+t^2}\,dt\\
&=2-2\,\Big[\arctan{t}\Big]_0^1\\
&=2-\frac{\pi}{2}\,.
\end{align*}

Hi Riemann. Hi Grigorios. Here is another solution.

\(\displaystyle{\begin{aligned} \int_{0}^{1}\dfrac{\sqrt{1-x^2}}{(1+x)^2}\,\mathrm{d}x&\stackrel{x=\sin\,u}{=}\int_{0}^{\pi/2}\dfrac{\sqrt{1-\sin^2\,u}\,\cos\,u}{(1+\sin\,u)^2}\,\mathrm{d}u\\&=\int_{0}^{\pi/2}\dfrac{\cos^2\,u}{(1+\sin\,u)^2}\,\mathrm{d}u\\&=\int_{0}^{\pi/2}\dfrac{(1-\sin\,u)(1+\sin\,u)}{(1+\sin\,u)^2}\,\mathrm{d}u\\&=\int_{0}^{\pi/2}\dfrac{1-\sin\,u}{1+\sin\,u}\,\mathrm{d}u\\&\stackrel{u=\pi/2-u'}{=}\int_{0}^{\pi/2}\dfrac{1-\cos\,u}{1+\cos\,u}\,\mathrm{d}u\\&=\int_{0}^{\pi/2}\dfrac{\sin^2\,(u/2)}{\cos^2(u/2)}\,\mathrm{d}u\\&=\int_{0}^{\pi/2}\tan^2\,(u/2)\,\mathrm{d}u\\&=2\,\int_{0}^{\pi/2}\dfrac{1}{2}\,(1+\tan^2\,(u/2))\,\mathrm{d}u-\int_{0}^{\pi/2}\mathrm{d}u\\&=2\,\left[\tan\,(u/2)\right]_{0}^{\pi/2}-\dfrac{\pi}{2}\\&=2-\dfrac{\pi}{2}\end{aligned}}\).