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"#
