Solving $\int_0^1 \int_0^x x \sqrt{x^2+3y^2} \,dy\, dx $ Solving $$\int_{0}^{1} \int_{0}^{x} x \sqrt{x^2+3y^2} \,dy\, dx $$
I tried doing this change of variable: $(x,y) = (u, \frac{v}{\sqrt{3}}) $
So the Jacobian is: $\frac{\sqrt{3}}{3}$, and the integral will be:
$$\int_0^1 \int_0^u u \sqrt{u^2+v^2} \frac{\sqrt{3}}{3} \,dv\, du $$
Then I tried using polar coordinates: $(u,v) = (r\cos(\theta),r\sin(\theta))$
And the integral became:
$$\frac{\sqrt{3}}{3}\int_0^{\pi/4} \int_0^{\sec(\theta)} r^2\cos(\theta) r \, dr \,d\theta = 
\frac{\sqrt{3}}{12}\int_0^{\frac{\pi}{4}} \sec^3(\theta) \, dr\, d\theta $$
But that won't give me the answer $$ \frac{\sqrt{3}}{24} ( 2 \sqrt3 + \ln(2+ \sqrt3)$$
Can someone help me?
Thanks.
Edit: Thanks for the answers. I want to understand also why my calculations are wrong. Can someone solve it using the change of variables that I used? Or are them not right?
 A: Note that
$$
\int_0^1 \int_{0}^{x} x \sqrt{x^2+3y^2} \,dy\, dx 
= \int_0^1 x \left(\int_{0}^{x} \sqrt{x^2+3y^2} \ dy\right) dx 
$$
and the inner integral can be taken using identity (29) at http://integral-table.com:
$$
\int \sqrt{a^2+y^2} \ dy
 = \frac{y}{2} \sqrt{a^2+y^2}
 + \frac{a^2}{2} \ln \left| y + \sqrt{y^2 + a^2} \right| + C
$$
A: Hint. One may integrate with respect to $y$ first performing the change of variable 
$$
y=\frac{x}{\sqrt{3}}\sinh t,\quad dy=\frac{x}{\sqrt{3}}\cosh t\:dt,
$$ giving, for $x \in [0,1]$,
$$
\begin{align}
\int_{0}^{x} x \sqrt{x^2+3y^2} dy&=\frac{x^3}{\sqrt{3}}\int_{0}^{\log\left(2+\sqrt{3}\right)} \sqrt{1+\sinh^2 t} \:\cosh t\:dt
\\\\&=\frac{x^3}{\sqrt{3}}\int_{0}^{\log\left(2+\sqrt{3}\right)}  \cosh^2 t\:dt
\\\\&=\frac{x^3}{2\sqrt{3}}\int_{0}^{\log\left(2+\sqrt{3}\right)}  (1+\cosh (2t))\:dt
\\\\&=\frac{x^3}{2\sqrt{3}}\left(2 \sqrt{3}+\log\left(2+\sqrt{3}\right)\right)
\end{align}
$$ then it is easier to integrate with respect to $x$.
A: let $u=\frac{\sqrt{3}y}{x}$ and $v=x$, so
$$\int_{0}^{1} \int_{0}^{x} x \sqrt{x^2+3y^2} \,dy\, dx=\frac{\sqrt{3}}{3}\int_{0}^{1} \int_{0}^{\sqrt{3}}v^3\sqrt{1+u^2}dudv=\frac{\sqrt{3}}{3}\int_{0}^{1}v^3dv \int_{0}^{\sqrt{3}}\sqrt{1+u^2}du $$
$$I=\frac{\sqrt{3}}{12}\frac{u\sqrt{1+u^2}+\ln(u+\sqrt{1+u^2})}{2}\mathcal{|_{0}^\sqrt{3}}=\frac{\sqrt{3}}{24}(2\sqrt{3}+\ln(2+\sqrt{3}))$$
