$$
\begin{align}
f(x) &= \sqrt{\frac{1+x}{1-x}}\\
f(-x)&= \sqrt{\frac{1-x}{1+x}}\\
f(x)+f(-x) &= \sqrt{\frac{1+x}{1-x}} + \sqrt{\frac{1-x}{1+x}}\\
&=\frac{2}{\sqrt{1-x^2}}\\
\int^{1}_{-1}f(x){\rm{d}}x &= \frac{1}{2}\int^{1}_{-1}(f(x)+f(-x)){\rm{d}}x\\
&=\int^{1}_{-1}\frac{1}{\sqrt{1-x^2}}{\rm{d}}x\\
&=\arcsin(1) - \arcsin(-1)\\& = \pi
\end{align}
$$
\begin{align}
f(x) &= \sqrt{\frac{1+x}{1-x}}\\
f(-x)&= \sqrt{\frac{1-x}{1+x}}\\
f(x)+f(-x) &= \sqrt{\frac{1+x}{1-x}} + \sqrt{\frac{1-x}{1+x}}\\
&=\frac{2}{\sqrt{1-x^2}}\\
\int^{1}_{-1}f(x){\rm{d}}x &= \frac{1}{2}\int^{1}_{-1}(f(x)+f(-x)){\rm{d}}x\\
&=\int^{1}_{-1}\frac{1}{\sqrt{1-x^2}}{\rm{d}}x\\
&=\arcsin(1) - \arcsin(-1)\\& = \pi
\end{align}
$$