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

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Answer 1 · 2017-02-23T19:33:01

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

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:

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

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

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

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

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

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