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I was given this integral: $$\int_{0}^1 \int_{0}^\infty \frac{y\arctan(xy)}{(1+y^2)(1+x^2y^2)} dy dx $$

I tried using Fubini to change the integration order and what I got was:

$$\int_{0}^\infty \frac{(\arctan(y^2) - \arctan(0))}{2(1+y^2)}dy $$ and I cannot integrate further, is this how I should have solved this integral?

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    what is the context? have you considered to do it numerically? – 2017-01-07
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    Note that $\arctan(0) = 0$, so you can "ignore" that part. – 2017-01-07
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    @Mark I do ignore it when I move to the next step. Just showed it as this was the process I did... – 2017-01-07
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    @user190080 It can be done numerically but as this is a home assignment I highly doubt this was my TA intent – 2017-01-07
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    There's a mistake in your integration with respect to $x$. It's not $\arctan(y^2)$, but $\arctan(y)^2$. And you should have lost the factor of $y$ in the numerator. After these corrections, the integral is straighforward. – 2017-01-07
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    @B.Goddard the y factor is my typo, correcting it now. THX!! – 2017-01-07

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The second integral should have the square outside the arctan. And $\arctan(0)$ is $0$, so your answer is $$ \frac{1}{2}\int_0^\infty \frac{\arctan(y)^2}{1+y^2}dy = \frac{1}{6}(\arctan(\infty)^3-\arctan(0)^3) = \frac{1}{6}\left(\frac{\pi}{2}\right)^3 $$