# Fastest way to integrate $\int_0^1 x^{2}\sqrt{x+x^2} \hspace{2mm}dx$

This integral looks simple, but it appears that its not so.

All Ideas are welcome, no Idea is bad, it may not work in this problem, but may be useful in some other case some other day ! :)

• Fastest way? Answer would always be to use a computer. Integral turns out to be $${{{{15} \ {\log \left( {17}{{-{{12} \ {\sqrt {2}}}}} \right)}}+{{244} \ {\sqrt {2}}}} \over {768} }\approx 0.380450122393$$ – gar Jul 8 '14 at 16:27

Note that $$x^2\sqrt{x+x^2}=\frac{x^3+x^4}{\sqrt{x(1+x)}}$$ So, let us look for a polynomial $P(x)=a x^3+bx^2+cx+d$ such that the derivative $(P(x)\sqrt{x(1+x)})^\prime$ is as close as we can to this function. An easy calculation shows that $$\left(P(x)\sqrt{x(1+x)}\right)^\prime=\frac{4 a x^4+(\frac{7 a }{2}+3 b) x^3+(\frac{5 b }{2}+2 c) x^2+(\frac{3 c }{2}+d )x+\frac{d}{2}}{\sqrt{x(1+x)}}$$ So, choosing $a=\frac{1}{4}$, $b=\frac{1}{24}$, $c=-\frac{5}{96}$ and $d=\frac{5}{64}$ we see that $$\left(P(x)\sqrt{x(1+x)}\right)^\prime=\frac{ x^4+ x^3}{\sqrt{x(1+x)}}+\frac{5}{64}\cdot\frac{1}{2\sqrt{x(1+x)}}$$ This reduces the considered integral to a simple one: $$\int_0^1x^2\sqrt{x(1+x)}dx=\Big[P(x)\sqrt{x(1+x)}\Big]_0^1-\frac{5}{64}\int_0^1\frac{dx}{2\sqrt{x(1+x)}}$$ The last integral is easy since $\log(\sqrt{x}+\sqrt{1+x})$ is a primitive of the integrand. Thus $$\int_0^1x^2\sqrt{x(1+x)}dx=\Big[P(x)\sqrt{x(1+x)}-\frac{5}{64}\log(\sqrt{x}+\sqrt{1+x})\Big]_0^1$$ Finally, $$\int_0^1x^2\sqrt{x(1+x)}dx= \frac{61}{96 \sqrt{2}}-\frac{5}{64} \log \left(1+\sqrt{2}\right).$$

HINT:

As $$x^2+x=\frac{(2x+1)^2-1^2}4$$ set $$2x+1=\sec\theta$$

For $x=0,\sec\theta=1\implies\theta=0$ and for $x=1,\sec\theta=3\implies \theta=\arccos\frac13$

• Hi, how to evaluate after that? Difficulty to evaluate appears to be the same! – gar Jul 8 '14 at 16:42
• @gar Check my work, but I think you get: $\sqrt{x+x^2}=\tan(\theta)/2$; $x^2=(\sec(\theta)-1)^2/4$; $dx=\sec(\theta) \tan(\theta)/2 d \theta$. So you need to be able to integrate $(\sec(\theta)-1)^2 (\sec(\theta)^2-1)/16$, which means you need to be able to integrate $\sec(\theta)^k$ for $k=0,1,2,3,4$. These are all fairly well known integrals. – Ian Jul 8 '14 at 23:24
• @Ian : I was unable to find your work, did you delete it? I always use CAS to do the well known integrals for me, and resort to pen and paper when it fails to get a closed form. – gar Jul 9 '14 at 3:25
• @gar I didn't write my work here, I was just working out Iab bhattacharjee's hint. – Ian Jul 9 '14 at 3:28

To calculate such integrals, it is often very useful change of variable of "type Euler". $$\sqrt{x+x^2} = x-t$$ with $x= \frac{t^2}{2t+1}$ and $dx= \frac{2t(t+1)}{2t+1}$

Integral reduces to an integral rational calculation:

$$\int_0^1 x^{2}\sqrt{x+x^2}dx =2\int_{1-\sqrt{2}}^0 \frac{t^6(t+1)^2}{(2t+1)^4}dt \cdots$$

• Nice one. But the denominator must be $(2t+1)^5$. – gar Jul 9 '14 at 3:18

HINT:

As $\displaystyle0\le x\le1,\sqrt{x(x+1)}=\sqrt x\sqrt{x+1}$

So, $$\int_0^1x^2\sqrt{x(x+1)}=\int_0^1x^{\dfrac52}\sqrt{x+1}\ dx$$

Set $x=\tan^2\theta$

Or integrating by parts, $$\int x^{\dfrac52}\sqrt{x+1}\ dx=x^{\dfrac52}\int\sqrt{x+1}\ dx-\int\left(\frac{d x^{\dfrac52}}{dx}\cdot\int\sqrt{x+1}\ dx\right)dx$$

$$=x^{\dfrac52}\cdot\frac{(x+1)^{\dfrac32}}{\dfrac32}-\dfrac52\int (x^2+x)^{\dfrac32}\ dx$$

Now for $x^2+x,$ we can use the substitution used in my other answer

$$\int_{0}^{1}x^{\frac{5}{2}}\sqrt{1+x}dx$$

Substitute $x = t^2$:

$$2\int_{0}^{1}t^6\sqrt{1+t^2}dt$$

Substitute $t = \sinh(u)$:

$$2\int_{0}^{\operatorname{arcsinh}(1)}\left[\sinh^8(u)-\sinh^6(u)\right]du$$

Let $x=\sinh^2 u$. (This is the same transformation as CountIblis used, but I'll employ it slightly differently.) Observe that $dx=2\cosh u \sinh u \, du$ and $$x+x^2=\sinh^2 u+\sinh^4 u=\sinh^2 u(1+\sinh^2 u)=\sinh^2 \cosh^2 u$$ since $\cosh^2 u-\sinh^2 u=1$.

Therefore \begin{align} \int_0^1 x^2 \sqrt{x+x^2}\,dx &=\int_0^{\sinh^{-1}1}\sinh^4 u\cdot \cosh u\sinh u \cdot \cosh u\sinh u\,du\\ &=\int_0^{\sinh^{-1} 1} \cosh^2 u \sinh^6 u\,du \end{align} To compute this integral, we recall that $\cosh u = \dfrac{1}{2}(e^u+e^{-u}),$ $\sinh u = \dfrac{1}{2}(e^u-e^{-u})$. One could expand the product in terms of exponentials and integrate term by term. For a bit less tedious route, first let $z=e^{-u}$ so that the integral takes the form

$$\int_{e^{-\sinh^{-1}(1)}}^{1} \frac{1}{2^8}\left(z+\frac{1}{z}\right)^2 \left(z-\frac{1}{z}\right)^6\frac{dz}{z}$$ We then expand this product into a sum of terms which can each be integrated by the power rule. The one obstacle is the weird-looking lower endpoint of $e^{-\sinh^{-1}1}$. However, one can show from $\sinh\sinh^{-1}(1)=1$ that this is equal to $\sqrt{2}-1$. That means that all that remains is the (admittedly rather tedious process) of integrating term-by-term and juggling square-roots.