How i can integrate this expression? I've read about integration, and i believe i understood concept correctly. But, unfortunately, the simplest exercise already got my stumbled. I need to find an integral of $x{\sqrt {x+x^2}}$. So i proceed as follows,
By the fundamental theorem of calculus:
$f(x)=\int[f'(x)]=\int[x\sqrt{x+x^2}]$,
First I've tried to apply chain rule and i end up with:  
$f'(x)=xu^\frac{1}{2}\frac{du}{2x+1}$ , not sure how i can proceed in this case.
Next I've tried to apply product rule:
If $f(x)=i(x)j(x)$, then $x\sqrt{x+x^2}=i'(x)j(x)+j'(x)i(x)$,
Using sum rule i could assume that, $f(x)=i(x)j(x)-\int[j'(x)i(x)]$,
Now, finding that $i'(x)=x, i(x)=\frac{x^2}{2}, j(x)=\sqrt{x+x^2}$  and $j'(x)=\frac{2x+1}{2\sqrt{x+x^2}}$, 
$f(x)$ should be of form $f(x)=\frac{x^2\sqrt{x+x^2}}{2}-\int\frac{x^2(2x+1)}{ 4\sqrt{x+x^2}}$, so now i should find integral of this fraction,
If i can assume, that $p(x)=\frac{a(x)}{b(x)}$, then $\frac{x^2(2x+1)}{ 4\sqrt{x+x^2}}=\frac{a'(x)b(x)-a(x)b'(x)}{(b(x))^2}$, hence:
$b(x)=2(x+x^2)^\frac{1}{4}, b'(x)=\frac{1}{2(x+x^2)^\frac{3}{4}}$ and as a result $a(x)$, should be $a(x)=\frac{(x+x^2)^\frac{3}{4}(4a'(x)(x+x^2)^\frac{1}{4}-4x^3-2x^2)}{2x+1}$, but now i don't now how substitute $a'(x)$, if i differentiate this expression i will get $a''(x)$. 
So my question is, what substitution i shall perform to obtain a(x), a'(x)?
Thank you! And forgive me my ignorance.      
 A: I hope you do not mind if I prefer to start from scratch. We have
$$ \int (2x+1)\sqrt{x^2+x}\,dx = C+\frac{2}{3}(x^2+x)^{3/2} \tag{1}$$
and the problem boils down to computing $\int\sqrt{x^2+x}\,dx$. Integration by parts gives
$$ \int \sqrt{x^2+x}\,dx = x\sqrt{x^2+x}-\int\frac{x+2x^2}{2\sqrt{x+x^2}}\,dx \tag{2}$$
hence
$$ 2\int\sqrt{x^2+x}\,dx = x\sqrt{x^2+x}+\frac{1}{2}\int \frac{x}{\sqrt{x^2+x}}\,dx \tag{3}$$
and the problem boils down to computing $\int\frac{x}{\sqrt{x^2+x}}$. With an argument similar to $(1)$, we have:
$$ \int\frac{2x+1}{\sqrt{x^2+x}}\,dx = C+2\sqrt{x^2+x}\tag{4} $$
so it is enough to compute $\int\frac{dx}{\sqrt{x^2+x}}$ and here it comes an interesting trick. Since
$$ \frac{1}{\sqrt{x^2+x}}=2\cdot\frac{\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}}{\sqrt{x}+\sqrt{x+1}}\tag{5}$$
we have
$$ \int\frac{dx}{\sqrt{x^2+x}}=2\log\left(\sqrt{x}+\sqrt{x+1}\right)\tag{6}$$
and by putting everything together
$$ \int x\sqrt{x^2+x}\,dx = \color{red}{C+\frac{x\left(8x^3+10x^2-x-3\right)}{24\sqrt{x+x^2}}+\frac{\log(\sqrt{x}+\sqrt{x+1})}{8}}.\tag{7}$$
It is not that simple, indeed!
A: Hint
$$x\sqrt{x + x^2} = x\sqrt{\left(x + \frac{1}{2}\right)^2 - \frac{1}{4}}$$
Then you may think about setting
$$x + \frac{1}{2} = y$$
Et cetera.
