Double integral problem (pretty basic)

chexmix
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Good day, all:

We recently hit double/triple integrals in my multivariable calculus course and I have found that my integration abilities are, well, *beyond* rusty ... and so the problem below, which is one of the very first on my current problem set, has me stumped.

Homework Statement



\int\int_{R}\frac{x}{1+xy} dA R = { [0,1]x[0,1] = {(x,y): 0 \leq x \leq1, 0 \leq y \leq1 }

The Attempt at a Solution



My first and nearly only step was to turn this into an iterated integral:

\int^{1}_{0}\int^{1}_{0}\frac{x}{1+xy} dx dy

... and this is where I begin to choke and sputter. I have started to try u-substitution on this with u = 1+xy, but didn't get anything that made sense to me; some hints I have found online seem to indicate that I should be able to perform "polynomial long division" to turn this into a sum or difference of two simpler integrals, but I guess I don't sufficiently understand polynomial long division to carry this out here.

Any hints would be much appreciated. I am considering dropping this course, but I would like to avoid that ... I need some serious integration mojo to be directly infused into my skull asap. :bugeye:

Regards,

Glenn
 
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Did you try integrating with respect to y first?
 
murmillo said:
Did you try integrating with respect to y first?

No. I'll check it out and see what happens. Thank you.

EDIT: That was key, of course. Jeez. ::strikes forehead::
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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