What is the antiderivative of #ln(x)^2#?

1 Answer
Feb 23, 2017

#int (lnx)^2dx = x(ln^2x -2lnx + 2 ) + C #

Explanation:

Integrate by parts, using #dx# as differential part:

#int (lnx)^2dx = xln^2x -int x d(ln^2x) #

#int (lnx)^2dx = xln^2x - 2int x lnx/xdx #

#int (lnx)^2dx = xln^2x - 2int lnxdx #

Integrating by parts again:

#int (lnx)^2dx = xln^2x -2 xlnx + 2 int x d(lnx) #

#int (lnx)^2dx = xln^2x -2 xlnx + 2 int x dx/x #

#int (lnx)^2dx = xln^2x -2 xlnx + 2 int dx #

#int (lnx)^2dx = xln^2x -2 xlnx + 2 x + C #

#int (lnx)^2dx = x(ln^2x -2lnx + 2 ) + C #