Can you help with this integration problem? Integration of $\displaystyle \int \dfrac{x}{x^2-\sqrt{x}} dx.$
I attempted substituting $\sqrt{x}$ with $u$ and then solving it using partial fractions but I'm not getting the right answer. I know I could used conjugates but I want to understand why the way I solved it at first was wrong. 
 A: You were by the right way, let me continue your work, if 
$$I=\displaystyle \int \dfrac{x}{x^2-\sqrt{x}} dx.$$
First you make $u=\sqrt{x}$, then $du=\dfrac{1}{2\sqrt{x}}dx$, so
$$I=2\displaystyle \int \dfrac{u^2}{u^3-1} du.$$
Then, you make the substitution $s=u^3-1$, so $ds=3u^2du$ and
$$I=\frac{2}{3}\displaystyle\int \dfrac{1}{s}ds = \frac{2}{3} \ln(s)+C $$
From that
$$I=\frac{2}{3} \ln(u^3-1)=\frac{2}{3} \ln(\sqrt{x}^3-1)+C$$
A: $$x^{1/2} = t,\\ 1/2x^{-1/2}dx=dt,\\ dx=2tdt$$
$$\int\frac{2t^3dt}{t^4-t}$$
Factor denominator
$$t(t^3-1)$$
You can eliminate $t$ from both numerator and denominator and what remains is to identify an internal derivative ( chain rule differential ) which you probably can do on your own.

Ok, seems I need to explain what I mean with "internal derivative" above. Consider the chain rule:
$$\frac{\partial f}{\partial t} = \frac{\partial f}{\partial g} \cdot \frac{\partial g}{\partial t}$$
In our case since we want to find an anti-derivative, we have $$\frac{\partial f}{\partial t} = \frac{2t^2}{t^3-1}$$ to try and factor in such a way.
if we choose $$\frac{\partial g}{\partial t} = kt^2, \frac{\partial f}{\partial g} = \frac{1}{t^3-1}$$
we can choose $g(t) = t^3-1$, then $g'(t) = 3t^2$ and so $k=3$
So doing substitution $g(t) = t^3-1$ gives $\frac{\partial f}{\partial g} = \frac 1 g$ and $f = \log(g)$
Now what remains is just to substitute back and simplify and we arrive at same solution as the other guys.
A: Another way, first note that
$$\int \frac{x}{x^2-\sqrt{x}}dx=\int \frac{x}{\sqrt{x}(x^{3/2}-1)}dx=\int \frac{x^{1/2}}{x^{3/2}-1}dx$$
Now, let $u=x^{3/2}-1$ and $du=\frac{3}{2}x^{1/2}dx$, then, the integral turns in
$$\int \frac{x^{1/2}}{x^{3/2}-1}dx= \frac{2}{3}\int \frac{1}{u}du=\frac{2}{3}\log (u)+c$$
Hence
$$\int \frac{x}{x^2-\sqrt{x}}dx=\frac{2}{3}\log (x^{3/2}-1)+c$$
A: $\sqrt{x} = u$ is fine, but I think $u^2 = x$ is slightly easier to work with.
$u^2 = x\\ 2u\ du = dx$
$\int \frac { u^2}{u^4 - u}(2u\ du)\\
\int \frac { 2u^2}{u^3 - 1}\ du$
And now we do a second substitution $v = u^3 - 1$
$\int \frac {2}{3v} \ dv = \frac 23 \ln v + c$
Going at it with your substitution should get us to the same place, it is just a little messier
$u = \sqrt x\\
du = \frac{1}{2\sqrt x} \ dx\\
\int \frac {2x^\frac 32}{(x^2 - \sqrt x)(2\sqrt x)}\ dx \\
\int \frac {2 u^3}{u^4 - u}\ dx$ 
etc.
