Question

help!

∫((ln x^2)/x)) dx

u= ln x^2
du= 2 ln x dx (?)

Not sure how to set this one up?

Answer 1

let u = ln(x^2) or 2lnx
then du/dx = 2/x
du = (2/x)dx

let dv = (1/x) dx
dv/dx = 1/x
v = lnx

then
∫ u dv = uv - ∫v du
∫2lnx(1/x) dx = 2lnx(lnx) - ∫lnx(2/x)dx
2 ∫2lnx / x dx = 2 (lnx)^2
∫(ln x^2)/x dx = (lnx)^2


proof:
d (lnx)^2) /dx = 2(lnx)(1/x) = ln (x^2) / x

Answer 2

Thank you!