TITLE: Integral asked on MIT bee qualifier 2012 $\int\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}}\,dx$
QUESTION [6 upvotes]: I have been trying to solve this one but I have no clue, I put it into wolfram and the result is absurd, considering this one is taken from MIT bee qualifier 2012, this is the integral:
$$\int\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} \, dx$$

REPLY [3 votes]: $$
\begin{align}
\int\frac{x-1}{x+1}\frac{\mathrm{d}x}{\sqrt{x^3+x^2+x}}
&=\int\frac{x^{-1/2}-x^{-3/2}}{x^{1/2}+x^{-1/2}}\frac{\mathrm{d}x}{\sqrt{\left(x^{1/2}+x^{-1/2}\right)^2-1}}\\
&=2\int\frac1{x^{1/2}+x^{-1/2}}\frac{\mathrm{d}\!\left(x^{1/2}+x^{-1/2}\right)}{\sqrt{\left(x^{1/2}+x^{-1/2}\right)^2-1}}\\
&=2\int\frac1u\frac{\mathrm{d}u}{\sqrt{u^2-1}}\\[9pt]
&=2\sec^{-1}(u)+C\\[15pt]
&=2\sec^{-1}\left(x^{1/2}+x^{-1/2}\right)+C
\end{align}
$$