I'm trying to solve the double integral $\displaystyle\int_0^1\int_0^1\dfrac{y}{x^2y^2+1}dx~dy$ . I'm guessing something with natural log will have to be done. Doing the steps of this problem are more important than the value itself.
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
0
I wouldn't substitute. Just integrate wrt $x$ first.
$$\newcommand{\d}[1][x]{\,\mathbb{d}#1} \int_0^1 \int_0^1\frac{y}{x^2y^2 + 1}\d[x] \d[y] = \int_0^1 \arctan{y} \d[y]$$
And this integral can be easily evaluated by parts ($\d[u] = 1$, $v = \arctan y$).
$\begingroup$
$\endgroup$
Make the substitution $\\u = x^{2}y^{2}+1, \\ du=(2x^{2}y)dy$
The rest is easy stuff.
