$ \int \frac{x^2-x}{x+\sqrt {x}}\, \mathrm{d}x$ My problem, is how to find the integral of $$ \int \frac{x^2-x}{x+\sqrt {x}}\, \mathrm{d}x$$
And I don't even know where to start. I tried factoring out x, but I don't get anything useful. I need some suggestions please. 
I appreciate your help!
 A: Observe that $x^2-x=(x+\sqrt{x})(x-\sqrt{x})$
A: $$ \int \frac{x^2-x}{x+\sqrt {x}}\, \mathrm{d}x= \int \frac{(x-1)\sqrt {x}}{1+\sqrt {x}}\, \mathrm{d}x=\int \left(\sqrt {x}-1\right)\sqrt {x}\, \mathrm{d}x$$
A: Let $t=\sqrt x$ then $dx=2tdt$ hence
$$ \int \frac{x^2-x}{x+\sqrt {x}}\, \mathrm{d}x=2\int\frac{t^4-t^2}{t+1}\, \mathrm{d}t=2\int t^2(t-1)\, \mathrm{d}t$$
which we can integrate easily.
A: We have $$ \int \frac{x^2-x}{x+\sqrt {x}}\, dx=\int (x-\sqrt{x} ) dx. $$
It is sum of table integrals.
A: Factor the top using $a^2-b^2=(a+b)(a-b)$:
$$ \int \frac{(x+\sqrt{x})(x-\sqrt{x})}{x+\sqrt {x}}\, \mathrm{d}x$$
Simplifying we obtain:
$$ \int x-\sqrt{x}, \mathrm{d}x$$
Which can be easily integrated to:
$$ \frac{x^2}{2}-\frac{2x^{3/2}}{3}+c$$
A: If you didn't see that the fraction simplifies, you can always integrate a rational expression involving only a single simple square root by changing variable to that square root. Here putting $u=\sqrt x$ gives $x=u^2$ so $\def\d{\,\mathrm d}\!\d x=2u\d u$, and you get
$$
  \int \frac{(x^2-x)\d x}{x+\sqrt {x}}=\int\frac{(u^4-u^2)2u\d u}{u^2+u}.
$$
Now you are forced to do the polynomial division and find that $2u(u^4-u^2)=2(u^2+u)(u^3-u^2)$, so you are looking at
$$
  \int2(u^3-u^2)\d u=\frac14u^4-\frac23u^3+C=\frac14x^2-\frac23x\sqrt x+C.
$$
