Indefinite integral with substitution For my engineering math course I got a couple of exercises about indefinite integrals. I ran trought all of them but stumbled upon the following problem. 
$$\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx $$
We can write $1+x-2x^2$ as $(1-x)(2x+1)$ 
So I got:
$$
\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx = \int \frac{1-x}{\sqrt{(1-x)(2x+1)}}\,dx 
$$
We can also replace $1-x$ in the denominator with $\sqrt{(1-x)^2}$ 
$$
\int \frac{1-x}{\sqrt{(1-x)(2x+1)}}\,dx = \int \frac{\sqrt{(1-x)^2}}{\sqrt{(1-x)(2x+1)}}\,dx
$$
If we simplify this fraction we get:
$$
\int \frac{\sqrt{1-x}}{\sqrt{2x+1}}\,dx
$$
Next we apply the following substitutions
$$
u = -x
$$
so : $-du = dx$
We can rewrite the integral as following:
$$-\int \frac{\sqrt{1+u}}{\sqrt{1-2u}}\,du$$
Then we apply another substitution: 
$\sqrt{1+u} = t $ so $ \frac{1}{2\sqrt{1+u}} = dt $ 
We rewrite: $ \sqrt{1+u} $ to $\frac{1}{2}t^2 \,dt $ 
We can also replace $\sqrt{1-2u} $ as following: 
$$\sqrt{-2t^2+3}=\sqrt{-2(1+u)+3}=\sqrt{1-2u}$$ 
With al these substitutions the integral has now the following form: 
$$-\frac{1}{2}\int \frac{t^2}{\sqrt{-2t^2+3}}\,dt$$ 
Next we try to ''clean'' up the numerator: 
$$-\frac{1}{2} \int \frac{t^2}{\sqrt{\frac{1}{2}(6-t^2)}} \, dt$$
$$-\frac{\sqrt{2}}{2} \int \frac{t^2}{\sqrt{6-t^2}} \, dt$$
And that's where I got stuck. I can clearly see that an arcsin is showing up in the integral but don't know how to get rid of the $t^2$.

Substitute $t = \sqrt{6} \sin(u)$, so $dt = \sqrt{6} \cos(u) du$ and the integrand becomes (up to losing the $-\frac{1}{\sqrt{2}}$ at the start)
$$\sqrt{6} \int \sqrt{6} \sin^2(u) du$$
I think you can do that!
