I have been trying to solve this double integral. I thought about converting the area in order to simplify the function inside the integral but I only used it to simplify the area of intagration, not the function itself so I don't know which conversion will help.
$$\int _0^3\int _0^3 (4y + x)\sqrt{x + y + 1} dxdy$$
Any help would be greatly appreciated!
If you do the change of variable $x=\frac13(-X+4Y)$ and $y=\frac13(X-Y)$, then $X=4x+y$ and $Y=x+y$, and your integral becomes$$\iint_R\frac13X\sqrt{Y+1}\,\mathrm dX\,\mathrm dY,\tag1$$where $R$ is the region of $\Bbb R^2$ bounded by the lines $Y=X(\iff y=0)$, $Y=X-9(\iff y=3)$, $Y=\frac X4(\iff x=0)$, and $Y=\frac{X+9}4(\iff x=3)$ (see the picture below); the $\frac13$ in $(1)$ is the absolute value of the Jacobian of$$(X,Y)\mapsto\left(\frac13(-X+4Y),\frac13(X-Y)\right).$$In fact, $R$ is the parallelogram whose vertices are $(0,0)$, $(3,3)$, $(12,3)$, and $(15,6)$. Therefore, $(1)$ is equal to on third of\begin{multline}\int_0^3\int_{X/4}^XX\sqrt{Y+1}\,\mathrm dY\,\mathrm dX+\int_3^{12}\int_{X/4}^{(X+9)/4}X\sqrt{Y+1}\,\mathrm dY\,\mathrm dX+\\+\int_{12}^{15}\int_{X-9}^{(X+9)/4}X\sqrt{Y+1}\,\mathrm dY\,\mathrm dX.\end{multline}In other words, it is equal to$$\frac{196\sqrt7}3-\frac{216}7.$$