TITLE: Integral $\int_{-1}^1 x \frac{\sqrt{1-x}}{\sqrt{1+x}}dx$
QUESTION [2 upvotes]: Right now I am trying to solve the following integral:
$$\int_{-1}^1 x \frac{\sqrt{1-x}}{\sqrt{1+x}}dx$$
So I have been trying to rationalize and simplify this since it looks to me that it will be reduced to trigonometric functions. I tried first to take the squared root from the nominator and separate the integral to a sum of the following two integrals:
$$\int_{-1}^1 \frac{{x}}{\sqrt{1-x^2}}dx + \int_{-1}^1 \frac{{-x^2}}{\sqrt{1-x^2}}dx$$
However, I am stuck at this part. Does anyone have any idea of other properties I can use to simplify this integral? The first one looks like can be solved by substitution but the second integral still needs to be rearranged. Thanks!

REPLY [0 votes]: By enforcing the substitution $\frac{1-x}{1+x}\mapsto z$ (notice that $f(x)=\frac{1-x}{1+x}$ is an involution) we get:
$$ \int_{-1}^{1}x\sqrt{\frac{1-x}{1+x}}\,dx=2\int_{0}^{+\infty}\frac{(1-z)\sqrt{z}}{(z+1)^3}\stackrel{z\mapsto u^2}{=}4\int_{0}^{+\infty}\frac{u^2-u^4}{(u^2+1)^3}\,du $$
and by integration by parts we get that the last integral equals
$$ \int_{0}^{+\infty}\frac{1-3u^2}{(1+u^2)^2}\,du =\color{blue}{-\frac{\pi}{2}}$$
since for any $a>0$ we have
$$ \int_{0}^{+\infty}\frac{du}{u^2+a}=\frac{\pi}{2\sqrt{a}}\quad\xrightarrow{\frac{d}{da}}\quad \int_{0}^{+\infty}\frac{du}{(u^2+a)^2}=\frac{\pi}{4a\sqrt{a}}.$$
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