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: ∫[-1,1]∫[-√(1 - y2),√(1 - y2)] ln(x2 + y2 + 1) Relevent Equations: x2 + y2 = r2 ∫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(r2 + 1)rdrdθ u = ln(r2 + 1) dv = rdr du = (2r)/(r2 + 1) dr v = 1/2 • r2 ∫[0,2π] 1/2 • r2 • ln(r2 + 1) |[0,1] -∫[0,2π]∫[0,2π] (r3)/(r2 + 1) • rdrdθ Note, I simplified vdu. 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 r4/(r2 + 1), which equals r2 - 1 + 1/(r2 + 1). So, the new integral is: - ∫[0,2π]∫[0,1] [r2 - 1 + 1/(r2 + 1)] drdθ The negative came from the earlier integration by parts step. I noticed that the third term in this integral fit the formula for tan-1. Integrating the second integral term by term and distributing the negative yields: ∫[0,2π] [-1/3 • r3 + r - tan-1(r)] |[0,1] dθ ∫[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: 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.