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 ! :)
 A: 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 $x=1,\sec\theta=3\implies \theta=\arccos\frac13$
A: 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,
$$\int_0^1x^2\sqrt{x(1+x)}dx=
\frac{61}{96 \sqrt{2}}-\frac{5}{64}
   \log \left(1+\sqrt{2}\right).
$$
A: 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:
$$\int_0^1 x^{2}\sqrt{x+x^2}dx =2\int_{1-\sqrt{2}}^0 \frac{t^6(t+1)^2}{(2t+1)^4}dt
\cdots $$
A: 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
A: $$\int_{0}^{1}x^{\frac{5}{2}}\sqrt{1+x}dx$$
Substitute $x = t^2$:
$$2\int_{0}^{1}t^6\sqrt{1+t^2}dt$$
Substitute $t = \sinh(u)$:
$$2\int_{0}^{\operatorname{arcsinh}(1)}\left[\sinh^8(u)-\sinh^6(u)\right]du$$
A: 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.
