Question
(ln(x))^2 dx, integrate.
Answers
Steve
use integration by parts. That is just the product rule in reverse.
d(uv) = u dv + v du
u dv = d(uv) - v du
∫u dv = ∫d(uv) - ∫v du
∫u dv = uv - ∫v du
So, here we just let
u = (lnx)^2
dv = dx
du = 2lnx * 1/x dx = (2lnx)/x
v = x
∫(lnx)^2 dx = x(lnx)^2 - ∫2lnx dx
Now repeat, this time letting
u = lnx
dv = 2dx
d(uv) = u dv + v du
u dv = d(uv) - v du
∫u dv = ∫d(uv) - ∫v du
∫u dv = uv - ∫v du
So, here we just let
u = (lnx)^2
dv = dx
du = 2lnx * 1/x dx = (2lnx)/x
v = x
∫(lnx)^2 dx = x(lnx)^2 - ∫2lnx dx
Now repeat, this time letting
u = lnx
dv = 2dx