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I'm on a tablet and having trouble with the math symbols so, for clarity, ∫[a,b] xdx is the integral from a to b of x with respect to x, and f(x) |[a,b] is a function of x evaluated from a to b.

Problem:

My Attempt:

Immediately I recognized that transforming this to a polar integral would make this much easier. The natural log is being integrated over a circular region radius 1 in the x-y plane, so the polar integral would be:

du = (2r)/(r

Evaluating the first integral yields:

Problem:

∫[-1,1]∫[-√(1 - y

Relevent Equations:^{2}),√(1 - y^{2})] ln(x^{2}+ y^{2}+ 1)

x

^{2}+ y^{2}= r^{2}∫udv = uv - ∫vdu

My Attempt:

Immediately I recognized that transforming this to a polar integral would make this much easier. The natural log is being integrated over a circular region radius 1 in the x-y plane, so the polar integral would be:

∫[0,2π]∫[0,1] ln(r

u = ln(r^{2}+ 1)rdrdθ

^{2}+ 1) dv = rdrdu = (2r)/(r

^{2}+ 1) dr v = 1/2 • r^{2}∫[0,2π] 1/2 • r

Note, I simplified vdu.^{2}• ln(r^{2}+ 1) |[0,1] -∫[0,2π]∫[0,2π] (r^{3})/(r^{2}+ 1) • rdrdθ

Evaluating the first integral yields:

∫[0,2π] 1/2 • ln2 dθ

The second integral is not in an inegrable form. To get it in integrable form, I performed long division. I can't really display that on here in a neat manor, but I just did r

^{4}/(r^{2}+ 1), which equals r^{2}- 1 + 1/(r^{2}+ 1). So, the new integral is:- ∫[0,2π]∫[0,1] [r

The negative came from the earlier integration by parts step. I noticed that the third term in this integral fit the formula for tan^{2}- 1 + 1/(r^{2}+ 1)] drdθ

^{-1}. Integrating the second integral term by term and distributing the negative yields:∫[0,2π] [-1/3 • r

∫[0,2π] [(-1/3 + 1 - π/4) - (0 + 0 - 0)] dθ

You can now either combine both integrals or evaluate them term by term. Once you evaluate, you get:^{3}+ r - tan^{-1}(r)] |[0,1] dθ∫[0,2π] [(-1/3 + 1 - π/4) - (0 + 0 - 0)] dθ

θ(1/2 • ln2 + 2/3 - π/4) |[0,2π]

2π(1/2 ln2 + 2/3 - π/4)

That's my final answer. The correct answer is π(ln4 - 1). These two are not equivalent.2π(1/2 ln2 + 2/3 - π/4)

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