I'm trying to solve this irrational integral $$ \int \frac{\sqrt{2x-x^2}+x}{2-\sqrt{2x-x^2}}\, dx$$
doing the substitution
$$ x= \frac{2t^2}{1+ t^2}$$
So $dx=\frac{4t}{(1+t^2)^2}dt$
and the integral becomes:
$$ \int \frac{4t^2(1+t)}{(1+t^2)^2(t^2-t+1)}\, dt$$
According to the bood ot should be
$$ \int \frac{4t^2(1+t)}{(1+t^2)^2(t-1)^2}\, dt$$
As usual I have checked my calculation but I don't understand where I'm making mistakes
Best Answer
HINT
Here I propose another way to solve it for the sake of curiosity.
I would start with noticing that
\begin{align*} \frac{\sqrt{2x - x^{2}} + x}{2 - \sqrt{2x -x^{2}}} & = -1 + \frac{2 + x}{2 - \sqrt{2x - x^{2}}}\\\\ & = -1 - \frac{1 - x}{2 - \sqrt{2x - x^{2}}} + \frac{3}{2 - \sqrt{2x - x^{2}}}\\\\ & = -1 - \frac{1 - x}{2 - \sqrt{2x - x^{2}}} + \frac{3}{2 - \sqrt{1 - (1-x)^{2}}} \end{align*}
To integrate the second term, you can make the substitution $u = 2x - x^{2}$.
To integrate the third term, you can make the substitution $\sin(\theta) = 1 - x$.
Can you take it from here?