Please help with natural log interals?

Question

(ln v^2 dv)
________
(ln e^v)

The high limit is e and the low limit is 1.

Answer 1

This one looks a lot more complex than it is. The key is to use the logarithm rules to simplify it.

∫(e,1) (ln v^2) / (ln e^v) dv

ln v^2 = 2 ln v
ln e^v = v ln e = v

∫(e, 1) 2 ((ln v) / v) dv
2 ∫(e, 1) ((ln v) / v) dv

Substitute:
u = ln v
du = 1/v dv

2 ∫(e, 1) u du

Integrate:
2∫u = u^2
u^2 | (e, 1)
(ln v)^2 | (e, 1)
ln e = 1, ln 1 =0
1^2 - 0 = 1 (solution!)