Question

Evaluate the integral: ∫xln|x+1|dx. ?

Answer 1

See below.

Explanation:

We must use integration by parts for this problem. We let `u = ln|x+ 1|` and `dv = x`. It follows that

`int u dv= uv - int vdu`

`int ln|x + 1|x dx= 1/2x^2ln|x + 1| - int 1/2x^2(1/(x + 1)) dx`

Let's consider the second integral, `int x^2/(x+ 1) dx` (we can multiply by a half at the end. Let `n = x + 1`. Then the integral becomes

`int (n - 1)^2/n dn = int (n^2 -2n + 1)/ndn = int n - 2 + 1/n dn = 1/2n^2 - 2n + ln|n| + C`

Summarizing:

`intxln|x +1| dx = 1/2x^2ln|x + 1| - 1/4n^2+ n - 1/2ln|n| + C`

`int xln|x + 1| dx = 1/2x^2ln|x + 1| - 1/4(x +1)^2 + x + 1 - 1/2ln|x +1| + C`

Hopefully this helps!