This is integral:
$\int \frac{x}{\sqrt[4]{x^3 (1-x)}}dx$
But I can write it like this:
So I can write $x$ as $\sqrt[4]{x^4}$
And I get:
$\int\sqrt[4]\frac{x}{1-x}dx$
Now I use substitution; Not sure how to solve it now,only thing I can do is set t=$\sqrt[4]\frac{x}{1-x}dx$
so i get $\frac{x}{1-x}=t^4$
$x=\frac{t^4}{1+t^4}$
$dx=4(x-1)^2\left(\frac{x}{1-x} \right )^{3/4}dt$
$(\frac{x}{1-x})^{3/4}$ is $t^3$ so $dx=4(x-1)^2t^3 dt $
Now I bring this back to $\int\sqrt[4]\frac{x}{1-x}dx$ Now I have:
$\int t^4 4(x^2-2x+1)dt$; so I use $x=\frac{t^4}{1+t^4}$ so change $x$;
now we have:
$4\int \left (\frac{t^4}{(1+t^4)^2}\right )dt$ so now I add +1-1; and we get 2 integrals:
$4(\int \frac{dt}{1+t^4}$-$\int \frac{dt}{(1+t^4)^2})$
So here I stucks,how do I solve these 2
Sorry for my bad English,hopefully you will understand the problem.