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. 

$$x^{1/2} = t,\\ 1/2x^{-1/2}dx=dt,\\ dx=2tdt$$
$$\int\frac{2t^3dt}{t^4-t}$$
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.
