I am half way stuck with this integral, any help would be appreciated.
Integral (p^5)ln(p) dp
u = ln p, du = 1/p
dv = p^5, v = p^6/6
Integral (p^5)ln(p) dp = p^6/6 ln(p) - Integral p^6/6p dp
Im stuck at the integral, substitution doesn't seem to work if I let w = p^6, dw = 6p^5..
Hello again, I stumbled with another difficult integral and need help. I didn't want to waste bandwidth so I added to this current thread.
I have integral e^2x sin3x dx
u = sin3x, du = 3cos3x
dv = e^2x, v = 1/2e^2x
integral e^2x sin3x dx = 1/2 e^2x sin3x - 3/2 integral e^2x cos3x dx
for integral e^2x cos3x dx
w = cos3x, dw = -3sin3x
dv = e^2x, v = 1/2e^2x
integral e^2x cos3x dx = 1/2e^2x cos3x + 3/2 integral e^2x sin3x dx
so combined is..
integral e^2x sin3x dx = 1/2 e^2x sin3x - 3/4 e^2x cos3x - 9/4 integral e^2x sin3x dx
since I have integral e^2x sin3x dx on both sides can I add integral e^2x sin3x dx making it
2 integral e^2x sin3x dx = 1/2 e^2x sin3x - 3/4 e^2x cos3x - 9/4
then divide both sides by 2
getting...
integral e^2x sin3x dx = [1/2 e^2x sin3x - 3/4 e^2x cos3x - 9/4]/2 + C?
Thanks for all your help.