What is the integral of lnx (natural log of x)?

Answer 1

xln(x) - x + C;

use parts:

integral of ln(x)
u = ln(x) du = 1/x dx
dv = 1 dx v = x

uv - (integral) v du
xlnx - (integral) x(1/x) dx

x(1/x) = 1

xlnx - (integral) 1 dx
= xlnx - x + C

Answer 2

must be done with integration by parts.

It will be x lnx - x + C

To solve:
Let u = ln x and dv = dx
then du = 1/x dx and v = x

uv - int (udv) = xlnx - int (x * 1/x)

= xlnx - int (1)
=xlnx - x + C

Answer 3

xln(x) - x

Derivative =
lnx + x/x -1 = lnx

Answer 4

integral ln x dx = xln x -x +C

Answer 5

∫ [ 1 / (x * ln(x)^2) ] dx <--- original integral u = ln(x) , du = (1 / x) dx <--- u-substitution ∫ [ 1 / (x * ln(x)^2) ] dx <--- original integral = ∫ [ 1 / u^2 ] du <--- substitute = ∫ [ u^(-2) ] du <--- rewrite = -u^(-1) <--- integrate = -1 / u <--- rewrite = -1 / ln(x) <--- substitute back now, plug in your limits of integration [ -1 / ln(4) ] - [ -1 / ln(2) ] = [ -1 / ln(4) ] + [ 1 / ln(2) ] = (1 / ln(2)) - (1 / ln(4)) = (ln(4) - ln(2)) / (ln(4) * ln(2)) = ln(4 / 2) / (ln(4) * ln(2)) = ln(2) / (ln(4) * ln(2)) = 1 / ln(4)