66Another way would be to set
1-x1+x = t2
which gives
x = 1-t21+t2 => dx = - 4t(1+t2)2 dt
The original integral now becomes
\int_0^1\dfrac{1-t^2}{1+t^2}\cdot t \cdot \dfrac{4t}{(1+t^2)^2}\ \mathrm dt
which, being a rational function, can be evaluated. However, it is quite lengthy. The easier one will be to set x = cos 2θ.
