How do you integrate $(ln x)^2 dx$ ?

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Answer 1 · 2016-05-22T00:27:49

Use integration by parts twice to find that

$intln^2(x)dx=xln^2(x)-2xln(x)+2x+C$

We will proceed using integration by parts :

Integration by parts (i):

Let $u = ln^2(x)$ and $dv = dx$
Then $du = 2ln(x)/xdx$ and $v = x$

Applying the integration by parts formula $intudv = uv - intvdu$

$intln^2(x)dx = xln^2(x) - int(2xln(x))/xdx$

$=xln^2(x)-2intln(x)dx" (1)"$

Integration by parts (ii):

Focusing on the remaining integral, let $u = ln(x)$ and $dv = dx$
Then $du = 1/xdx$ and $v = x$

Applying the formula:

$intln(x)dx = xln(x) - intx/xdx$

$=xln(x) - intdx$

$=xln(x) - x + C$

Substituting this back into $"(1)"$ :

$intln^2(x)dx = xln^2(x)-2(xln(x)-x)+C$

$=xln^2(x)-2xln(x)+2x+C$

How do you integrate (ln x)^2 dx(ln x)^2 dx?