I need to compute the following integral: $$\int \sqrt[4]{x(1+x)^3} dx$$
This problem is in a worksheet to exercise before a test and I can't solve it. In the worksheet the author suggest the sustitution $x=\frac{1}{t}$. After doing that I don't know how to proceed. I put this integral in Wolfram Alpha and the solution offered there is pretty much "ugly", so I think there is a nice answer as I did all the others integrals in a pretty clean way.
The other thing that may have happened is that there may be a typo in the worksheet and perhaps the author wanted to put another integral very similar to this, but the probability that this is the case I think is very low.
Any hints are appreciated.
Let's take the hint and run with it. Let $x = \frac{1}{t}$ then the integral becomes:
$$\int -\sqrt[4]{\frac{1}{t}\left(1+\frac{1}{t}\right)}\frac{1}{t^2}dt = \int -\frac{\sqrt[4]{(1+t)^3}}{t^3}dt$$
taking $x>0$ (there is another branch where $x<-1$ but we won't worry about this now). Now let $u^4 = 1+t$:
$$= \int - \frac{4u^6}{(u^4-1)^3}du = \int -4\left(\frac{u^2}{u^4-1}\right)^3du$$
this form allows us to do partial fraction decomposition more easily by symmetry arguments instead of having to solve a horrendous system of equations. Using $(a+b)^3 = a^3 + 3ab(a+b) + b^3$:
$$ = -\frac{1}{2} \int \left(\frac{1}{u^2-1} + \frac{1}{u^2+1}\right)^3du$$ $$ = -\frac{1}{2}\int \frac{1}{(u^2-1)^3} + 3\left(\frac{1}{u^2-1}\right)\left(\frac{1}{u^2+1}\right)\left(\frac{1}{u^2-1} + \frac{1}{u^2+1}\right) + \frac{1}{(u^2+1)^3}du$$
$$ = -\frac{1}{2} \int \frac{1}{(u^2-1)^3} + \frac{3}{2}\left(\frac{1}{(u^2-1)^2} - \frac{1}{(u^2+1)^2}\right) + \frac{1}{(u^2+1)^3}du$$
$$=-\frac{1}{4}\Biggr[\int \frac{1}{(u^2-1)^2}\left(\frac{2}{u^2-1}+3\right)du + \int \frac{1}{(u^2+1)^2}\left(\frac{2}{u^2+1}-3\right)du \Biggr]$$
Now, for the integral on the right use $u=\tan\theta$, which simplifies the expression to
$$\frac{1}{4}\int \frac{3}{4} - \frac{1}{4}\cos(4\theta)+\frac{1}{2}\cos(2\theta)d\theta = \frac{3}{16}\theta -\frac{1}{64}\sin(4\theta) + \frac{1}{16}\sin(2\theta)$$ $$= \frac{3}{16}\theta + \frac{1}{16}\sin\theta\cos\theta(2\sin^2\theta+1) = \frac{3}{16}\tan^{-1}(u) + \frac{1}{16}\frac{u}{1+u^2}\left(\frac{2u^2}{1+u^2}+1\right)$$
The integral on the left will need more partial fraction decomposition:
$$-\frac{1}{16}\int \left(\frac{1}{u-1}-\frac{1}{u+1}\right)^3+3\left(\frac{1}{u-1}-\frac{1}{u+1}\right)^2du$$
$$=-\frac{1}{16}\int \frac{1}{(u-1)^3} - 3 \left(\frac{1}{u-1}\right)\left(\frac{1}{u+1}\right)\left(\frac{1}{u-1}-\frac{1}{u+1}\right) - \frac{1}{(u+1)^3} +3\left(\frac{1}{u-1}-\frac{1}{u+1}\right)^2 du$$
$$=-\frac{1}{16}\int \frac{1}{(u-1)^3} - \frac{1}{(u+1)^3} +\frac{3}{2}\left(\frac{1}{u-1}-\frac{1}{u+1}\right)^2 du$$
$$=-\frac{1}{16}\int \frac{1}{(u-1)^3} - \frac{1}{(u+1)^3} +\frac{3}{2}\frac{1}{(u-1)^2} + \frac{3}{2}\frac{1}{u-1}- \frac{3}{2}\frac{1}{u+1} + \frac{3}{2}\frac{1}{(u+1)^2} du$$
$$= \frac{1}{32}\left(\frac{1}{(u-1)^2}-\frac{1}{(u+1)^2}\right) + \frac{3}{32}\left(\frac{1}{u-1}-\frac{1}{u+1}\right)+\frac{3}{32}\log\left(\frac{u+1}{u-1}\right)$$
From here, just add the answers of the two parts and plug in $u = \sqrt[4]{1+\frac{1}{x}}$.