Take the integral:
int x log(x + 1) dx
For the integrand x ln(x + 1), substitute u = x + 1 and du = dx:
= int(u - 1) ln(u) du
For the integrand (u - 1) log(u), integrate by parts, int f dg = f g - int g df where
f = ln(u), dg = (u - 1) du, df = 1/u du, g = 1/2 (u - 1)^2:
= (ln(u))/2 - u ln(u) + 1/2 u^2 ln(u) - 1/2 int(u - 1)^2/u du
For the integrand (u - 1)^2/u, substitute s = u - 1 and ds = du:
= (ln(u))/2 - u ln(u) + 1/2 u^2 ln(u) - 1/2 int s^2/(s + 1) ds
For the integrand s^2/(s + 1), do long division:
= (ln(u))/2 - u ln(u) + 1/2 u^2 ln(u) - 1/2 integral(s + 1/(s + 1) - 1) ds
Integrate the sum term by term and factor out constants:
= (ln(u))/2 - u ln(u) + 1/2 u^2 ln(u) - 1/2 int/(s + 1) ds - 1/2 int s ds + 1/2 int ds
For the integrand 1/(s + 1), substitute p = s + 1 and dp = ds:
= (ln(u))/2 - u ln(u) + 1/2 u^2 ln(u) - 1/2 int1/p dp - 1/2 int s ds + 1/2 int1 ds
The integral of 1/p is log(p):
= -(ln(p))/2 + (ln(u))/2 - u ln(u) + 1/2 u^2 ln(u) - 1/2 int s ds + 1/2 int1 ds
The integral of s is s^2/2:
= -s^2/4 - (ln(p))/2 + (ln(u))/2 - u ln(u) + 1/2 u^2 ln(u) + 1/2 int1 ds
The integral of 1 is s:
= -(ln(p))/2 - s^2/4 + s/2 + 1/2 u^2 ln(u) - u ln(u) + (ln(u))/2 + "constant"
Substitute back for p = s + 1:
= -s^2/4 + s/2 - 1/2 ln(s + 1) + 1/2 u^2 ln(u) - u ln(u) + (ln(u))/2 + "constant"
Substitute back for s = u - 1:
= 1/2 u^2 ln(u) - 1/4 (u - 1)^2 + (u - 1)/2 - u ln(u) + "constant"
Substitute back for u = x + 1:
= -x^2/4 + x/2 + 1/2 (x + 1)^2 ln(x + 1) - (x + 1) ln(x + 1) + "constant"
Which is equal to:
Answer:
1/4 (2 (x^2 - 1) ln(x + 1) - (x - 2) x) + "constant"
