Integral Question!?

Question

Can someone help compute this integral:

f x ln(2x-1) dx

Answer 1

Integral ( x ln(2x - 1) dx)

To solve this, we use integration by parts.

Let u = ln(2x - 1). dv = x dx
du = [1/(2x - 1)](2) dx. v = (1/2)x^2

Remember that parts work like this:

uv - Integral (v du), so we have

(1/2)x^2 ln(2x - 1) - Integral ( (1/2)x^2 [1/(2x - 1)](2) dx )

Multiply the constants in the integral.

(1/2)x^2 ln(2x - 1) - Integral ( x^2 [1/(2x - 1)] ) dx

Merge into one fraction within the integral.

(1/2)x^2 ln(2x - 1) - Integral ( x^2/(2x - 1) ) dx

Now, we have to do long division to reduce x^2 / (2x - 1) into a more integrable form. Without showing you the details,
x^2 / (2x - 1) = (1/4)/(2x - 1) + (1/2)x + 1/4, so we have

(1/2)x^2 ln(2x - 1) - Integral ( (1/4)/(2x - 1) + (1/2)x + (1/4) ) dx

This is much easier to solve now. Note that the integral of
(2x - 1) will be ln |2x - 1| offsetted by (1/2) due to the chain rule.

(1/2)x^2 ln(2x - 1) - [ (1/4)(1/2) ln|2x - 1| + (1/2)(1/2)x^2 + (1/4)x + C

Simplifying this,

(1/2)x^2 ln(2x - 1) - [ (1/8) ln|2x - 1| + (1/4)x^2 + (1/4)x ] + C

(1/2)x^2 ln(2x - 1) - (1/8) ln|2x - 1| - (1/4)x^2 - (1/4)x + C

Answer 2

use integration by part