What is the integral of xlnx - x (second integral of lnx).

∫ x(ln(x) - 1) dx

u = x
du = dx

dv = ln(x) - 1
v = x(ln(x) - 1) - x

x²(ln(x) - 1) - x² - ∫ x(ln(x) - 1) - x dx
x²(ln(x) - 2) - ∫ x(ln(x) - 1) dx + ∫ x dx
x²(ln(x) - 2) + (x²/2) - ∫ x(ln(x) - 1) dx
x²(ln(x) - (3/2)) - ∫ x(ln(x) - 1) dx

The remaining integral is the same as the original, so we can add it to both sides and divide by 2 to give.
(x²/2)(ln(x) - (3/2))



In general, the nth integral of ln(x) is
(xⁿ/n!)(ln(x) - H(n))

where H(n) is the nth harmonic number given by ∑{i = 1, n} 1/i.

∫(xlnx - x)dx = ∫x*ln(x) dx - ∫x dx

The second integral is easy, so just subtract it from the first integral. For the first integral, integrate by parts...

u = ln(x).........v' = x
u' = 1/x..........v = x²/2

So...

∫(xlnx - x)dx = [x²ln(x)]/2 - ∫(1/x)(x²/2)dx - x²/2 + C
= [x²ln(x)]/2 - (1/2)∫(x)dx - x²/2 + C
= [x²ln(x)]/2 - (1/2)(x²/2) - x²/2 + C
= [x²ln(x)]/2 - (1/4)(x²) - x²/2 + C
= (x²/2)[ln(x) - 1/2 - 1] + C
= (x²/2)[ln(x) - 3/2] + C