<s>[INST] Evaluate $\int\frac{x{\rm d}x}{(1+x)\sqrt{1-x-x^2}}$ My solution
Make a Euler's substitution. Let $\sqrt{1-x-x^2}=xz-1$. Then
\begin{align*}z&=\frac{1+\sqrt{1-x-x^2}}{x},\\x&=\frac{2z-1}{z^2+1},\\
{\rm d}x&=\frac{2(1+z-z^2)}{(z^2+1)^2}{\rm d}z,\\(1+x)\sqrt{1-x-x^2}&=\frac{z(z+2)(1+z-z^2)}{(z^2+1)^2}.
\end{align*}
Thus
\begin{align*}
\int\frac{x{\rm d}x}{(1+x)\sqrt{1-x-x^2}}&=2\int\frac{2z-1}{z(z^2+1)(z+2)}{\rm d}z\\
&=2\int\left(\frac{1}{z^2+1}+\frac{1}{2(z+2)}-\frac{1}{2z}\right){\rm d}z\\
&=2\arctan z+\ln|z+2|-\ln|z|+C\\
&=2\arctan z+\ln\left|\frac{z+2}{z}\right|+C.
\end{align*}
The result is right? Who can verify it for me? Thanks in advance.
 [/INST] If you want an alternative method,
as $1-x-x^2=\dfrac{5-(2x+1)^2}4,$ let $2x+1=\sqrt5\sin t$
$$\int\dfrac x{(1+x)\sqrt{1-x-x^2}}\ dx=\int\dfrac{\sqrt5\sin t-1}{\sqrt5\sin t+1}\ dt$$
$\dfrac{\sqrt5\sin t-1}{\sqrt5\sin t+1}=1-\dfrac2{\sqrt5\sin t+1}$
For the second part, use Weierstrass substitution
</s>