# Double integral problem My idea was that the limits are and that the anti-derivative of dy was

xlog(1+y^2)

but that seems wrong... gives us then we take dy guess, i figured it out eventually with the help of wolfram with the last integration

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SammyS
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The reason your 1st attempt didn't work is that $$\int\frac{1}{1+y^2}\,dy=\tan^{-1}(y)+C\,.$$

HallsofIvy
Homework Helper My idea was that the limits are and that the anti-derivative of dy was

xlog(1+y^2)

but that seems wrong...
Yes, it is wrong. It is standard "Calculus I" mistake to treat a function of the variable as if it were just the variable but you should be past that by the time you are doing multiple integrals. "$1/(1+ y^2)$" is NOT the same as $1/y$ and its anti-derivative is not a logarithm. The anti-derivative of $1/(1+ y^2)$ is $arctan(y)+ C$. That's a standard anti-derivative that you should have memorized. gives us then we take dy guess, i figured it out eventually with the help of wolfram with the last integration
Yes, reversing the order of integration is the best way to handle this one. Integerating $arctan(1)- arctan(x^2)= \pi/4- arctan(x^2)$ is likely to be very difficult!

SammyS
Staff Emeritus
Yes, reversing the order of integration is the best way to handle this one. Integerating $arctan(1)- arctan(x^2)= \pi/4- arctan(x^2)$ is likely to be very difficult!