integration by parts?

Question

integral = z(ln z)^2 dz

I am supposed to use substitution and integration by parts to solve this problem. I guess u = ln z and du = 1/z dz
But How do I use integration part when I only have U and don't have V?

Answer 1

You should use U = (ln z)^2 and dV = z dz, thus, dU = 2(ln z)/z dz and V = z^2/2.

Answer 2

Integral (z [ln(z)]^2 dz)

Use integration by parts.

Let u = [ln(z)]^2. dv = z dz
du = 2ln(z) (1/z) dz. v = (1/2)z^2

(1/2)z^2 [ln(z)]^2 - Integral ( (1/2)z^2 (2)ln(z) (1/z) dz )

(1/2)z^2 [ln(z)]^2 - Integral (z ln(z) dz)

Use integration by parts one more time.

let u = ln(z). dv = z dz.
du = (1/z) dz. v = (1/2)z^2

(1/2)z^2 [ln(z)]^2 - [(1/2)z^2 ln(z) - Integral ( (1/z)(1/2)z^2 dz ) ]

(1/2)z^2 [ln(z)]^2 - (1/2)z^2 ln(z) + Integral ( (1/z)(1/2)z^2 dz )

(1/2)z^2 [ln(z)]^2 - (1/2)z^2 ln(z) + Integral ( (1/2)z dz )

(1/2)z^2 [ln(z)]^2 - (1/2)z^2 ln(z) + (1/2)(1/2)z^2 + C

(1/2)z^2 [ln(z)]^2 - (1/2)z^2 ln(z) + (1/4)z^2 + C

Answer 3

You can make the substitution y = lnz, then the integral becomes:

∫ y² exp(2y) dy

There's your substitution. Then you integrate by parts twice. The first time:

u = y² and v = ½ exp(2y) that gives you:

uv - ∫ v du = ½ y² exp(2y) - ∫ y exp(2y) dy

then integrate by parts again using

u = y and v = ½ exp(2y)

that gives you ½ y exp(2y) - ½ ∫ exp(2y) dy

then you can evaluate the 3rd integral to get ¼ exp(2y)

½ y² exp(2y) - ½ y exp(2y) + ¼ exp(2y).

Now you can substitute for y (y = ln z) to get the answer in terms of z.

You should check to make sure that I didn't make any careless mistakes. I'm tired.