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:


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.


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!

  • $\begingroup$ Maybe a word to explain the downvote? Is it asking too much? $\endgroup$ – Jack D'Aurizio Sep 24 '16 at 23:14


$$x\sqrt{x + x^2} = x\sqrt{\left(x + \frac{1}{2}\right)^2 - \frac{1}{4}}$$

Then you may think about setting

$$x + \frac{1}{2} = y$$

Et cetera.

  • 1
    $\begingroup$ Is it really a hint that leads to somewhere? $\endgroup$ – Jack D'Aurizio Sep 24 '16 at 12:35
  • $\begingroup$ @JackD'Aurizio After the substitutions you have $$\int \left(y - \frac{1}{2}\right)\sqrt{y^2 - \frac{1}{4}}$$ The first term is trivial, and gives you immediately $\frac{1}{3}\left(y^2 - \frac{1}{4}\right)^{3/2}$. The other one might be a bit more "difficult" but it leads to a log form in quite straightforward way. $\endgroup$ – Turing Sep 24 '16 at 12:40
  • 1
    $\begingroup$ I think that to find the logarithmic (or $\text{arcsinh}$) part of the primitive is the crucial part of the exercise, so this is just a very partial hint to the solution. $\endgroup$ – Jack D'Aurizio Sep 24 '16 at 12:44
  • 3
    $\begingroup$ @JackD'Aurizio Well we are not here to solve other people's homework, as LOTS of other users used to tell me, many times. SO I started giving hints, or solving exercises when they may be really hard. I don't know what the OP knows and what is his level of study, but this integral is really one of the easiest ones for me. (Thanks my dear Calculus professor A.K., may long he live!) $\endgroup$ – Turing Sep 24 '16 at 12:48

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