Evaluate the integral from 25 to 36 of ln(y)/sqrt(y) dy?

Question

i did integration by parts and ended up with 144ln(36) -250/3ln(25)-2/3

and that was not right so if you could help me that would be sooooo helpful!

thanks

Answer 1

∫ (lny /√y) dy =

let √y = t →

y = t² →

dy = 2t dt

thus, substituting, you get:

∫ (lny /√y) dy = ∫ [ln(t²) / t] 2t dt =

t canceling out,

∫ 2 ln(t²) dt =

that is, owing to log properties,

∫ 2 (2 ln t) dt =

4 ∫ ln t dt =

now, integrating by parts, let:

ln t = u → (1/ t) dt = du

dt = dv → t = v

thus:

4 ∫ ln t dt = 4 [t ln t - ∫ t (1/ t) dt] =

4t ln t - 4 ∫ dt =

4t ln t - 4t + C

then, substituting back t = √y, you get the antiderivative:

∫ (lny /√y) dy = 4√y ln(√y) - 4√y + C

then, evaluating the definite integral from 25 to 36, you get:

[4√36 ln(√36) - 4√36] - [4√25 ln(√25) - 4√25] =

[4(6) ln(6) - 4(6)] - [4(5) ln(5) - 4(5)] =

24 ln(6) - 24 - 20 ln(5) + 20 =

24 ln(6) - 20 ln(5) - 4 = (approx.) 6.8134

I hope this helps...

Bye!