Question

How does one integrate the following:

Answer 1

Consider substituting $u=4+xy$, so that $\mathrm du=x\,\mathrm dy$. Then

$\displaystyle\int_{x=a}^{x=b}\int_{y=a}^{y=b}\frac x{(4+xy)^2}\,\mathrm dy\,\mathrm dx=\int_{x=a}^{x=b}\int_{u=4+ax}^{u=4+bx}\frac1{u^2}\,\mathrm du\,\mathrm dx$
$=\displaystyle\int_{x=a}^{x=b}\left(-\frac1u\right)\bigg|_{u=4+ax}^{u=4+bx}\,\mathrm dx$
$=\displaystyle\int_{x=a}^{x=b}\left(\frac1{4+ax}-\frac1{4+bx}\right)\,\mathrm dx$

Then by similar substitutions, you can easily find that you end up with

$\dfrac1a\ln|4+ax|-\dfrac1b\ln|4+bx|\bigg|_{x=a}^{x=b}$
$=\dfrac1a\ln|4+ab|-\dfrac1b\ln|4+b^2|-\dfrac1a\ln|4+a^2|+\dfrac1b\ln|4+ab|$
$=\dfrac1a\ln\left|\dfrac{4+ab}{4+a^2}\right|+\dfrac1b\ln\left|\dfrac{4+ab}{4+b^2}\right|$

Of course, this all assumes that the integrand is continuous over the domain of integration, which would require that $a,b$ are chosen such that $xy\neq-4$ for any $(x,y)\in[a,b]^2$. If in particular $ab>-4$, then we can write

$=\dfrac1a\ln\dfrac{4+ab}{4+a^2}+\dfrac1b\ln\dfrac{4+ab}{4+b^2}$

and you can combine the logarithms if you like as

$=\ln\sqrt[a]{\dfrac{4+ab}{4+a^2}}+\ln\sqrt[b]{\dfrac{4+ab}{4+b^2}}$
$=\ln\left(\sqrt[a]{\dfrac{4+ab}{4+a^2}}\sqrt[b]{\dfrac{4+ab}{4+b^2}}\right)$