# Polynomial long division two variables: $\frac{7 x}{1+7 xy}$

So I'm doing this iterated integral $\displaystyle $$\int_1^4\int_0^5$$ \dfrac{7 x}{1+7 xy} \, dx dy$

But I don't know how to rewrite the fraction.

If it was just $\dfrac{x}{1+7 x}$, then it'd have been rewritten as: $\frac{1}{7} - \frac{1}{7(7x+1)}$

But how do I start doing long division I end up with:

7xy + 1| x

then what? y is treated as constant...

if I start with 1 it'll be:

7xy+1 | x

     |7xy + 1


x - 7xy ... improbable...

if you didn't get it, just try doing the long division, divide x by 7xy+1

• No need to do long division. Just switch the order of integration. Then you can use $u$-substitution. – Tim Raczkowski Mar 9 '16 at 0:41
• If $y$ is (temporarily) constant, $7xy+1$ goes into $x$ $\dfrac{1}{7y}$ times. – Steve Kass Mar 9 '16 at 0:42
• @TimRaczkowski how do you do long division with x/(7xy+1) though? – Jack Mar 9 '16 at 23:17

It's not "really" two variables because you will be integrating with respect to one variable at a time: the way you have written it, first $x$ then $y$. So $$\frac{7x}{1+7xy}=\frac1y\frac{7xy}{1+7xy}=\frac1y\frac{1+7xy-1}{1+7xy} =\frac1y\Bigl(1-\frac1{1+7xy}\Bigr)\ .$$
Use Fubini: it is $$\int_1^4\biggl(\int_0^5\frac{7x}{1+7xy}\,\mathrm d\mkern 1mu y\biggr)\mathrm d\mkern 1mu x=\int_1^4(\ln(1+7xy)\bigg\vert_{y=0}^{y=5}\mathrm d\mkern 1mu x=\int_1^4\ln(1+35x)\mathrm d\mkern 1mu x.$$ Can you take it from here?