Question

How does one integrate the following:


`\int\limits^a_b \int\limits^a_b { (x)/((4+xy)^(2) ) } \, dy \, dx`

you can switch the bounds however you want (integrate with respect to x before y). I would like to know the best way to handle this problem.

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


`\frac1a\ln|4+ax|-\frac1b\ln|4+bx|\bigg|_(x=a)^(x=b)`

`=\frac1a\ln|4+ab|-\frac1b\ln|4+b^2|-\frac1a\ln|4+a^2|+\frac1b\ln|4+ab|`

`=\frac1a\ln\left|(4+ab)/(4+a^2)\right|+\frac1b\ln\left|(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\\eq-4` for any
`(x,y)\in[a,b]^2`. If in particular
`ab>-4`, then we can write


`=\frac1a\ln(4+ab)/(4+a^2)+\frac1b\ln(4+ab)/(4+b^2)`

and you can combine the logarithms if you like as


`=\ln\sqrt[a]{(4+ab)/(4+a^2)}+\ln\sqrt[b]{(4+ab)/(4+b^2)}`

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