Recent content by iismitch55

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}...

I'm going to also add in a picture of mywork. It may help since we are having trouble with latex.

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

My fault. I wrote the initial problem backwards. Should be x = +/- √(1 - y2). I will edit it so no one gets confused. I did draw the circle though.

100% correct. Sorry for the messyness. My tablet doesn't like to do the limits, and I'm not that experienced with how the code works.

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) relevant Equations: x2 + y2 = r2 ∫udv =...

Can confirm, tried, doesn't work. Thank you for all your help!

Thanks, I will give this a shot. Could you tell me why the denominator breaks up this way rather than (s + 1) each denominator?

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...

23.) y'' + 2y' + y = 4e-t; y(0) = 2, y'(0) = -1 Y(s) = [(as + b) y(0) + a y'(0) + F(s)]/(as2 + bs + c) My attempt: a = 1, b = 2, c = 1 F(s) = 4 L{ e-t } = 4/(s+1) (From Laplace Transform Table) Plugging and simplifying: Y(s) = (2s2 + 5s + 7)/[(s + 1)(s2 + 2s + 1) Here is where I get...