I didn't remember the antiderivative of lnu du off hand. I reworked the problem today. The mistake I made was in this step:
$$\frac{1}{2} \ln(r^2 + 1) - \int \frac{2r • r^2}{2(r^2 + 1)} r dr$$
I stuck an extra r in next to dr. I assumed that if I integrated again in polar (drdθ) then I needed...
Sorry, fell asleep last night.
So, the result of the indefinite integral using integration by parts:
$$\int r\ln(r^2 + 1)dr$$
$$u = \ln(r^2 + 1); dv = rdr$$
$$du = \frac{2r}{r^2 + 1} dr; v = \frac{1}{2} r^2$$
$$\frac{1}{2} r^2 \ln(r^2 + 1) - \int \frac{2r • r^2}{2(r^2 + 1)} rdr$$
$$\frac{1}{2}...
Okay, I will give latex a shot. Are you looking for the final indefinite integral with respect to r?
That would be:
\frac{1}{2} ln2 + \frac{2}{3} - \frac{π}{4}
If you would like, I can try retyping it with latex, although I cannot seem to get it to work properly
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 =...
So, which would be the correct way to proceed with partial fractions:
A(s + 1)^2(s + 1)^3 + B(s +1)(s + 1)^3 + C(s + 1)(s + 1)^2 = 2s^2 + 5s + 7
Or
A(s + 1)(s + 1) + B(s +1)(s + 1) + C(s + 1)(s + 1) = 2s^2 + 5s + 7
My PFD is a bit rusty sorry. Method 2 is what I tried before, which yielded...