Question

what is the integral of (lnx)^2 dx

Answer 1

Try substituting u=ln(x).

Answer 2

Let u = [ ln(x) ]^2. dv = dx.
du = 2ln(x) (1/x) dx. v = x

Recall the formula of integration by parts:

uv - Integral ( v du )

x [ln(x)]^2 - Integral ( x (2 ln(x)) (1/x) dx )

And we get some cancellations in the integral, particular with x and (1/x).

x [ln(x)]^2 - Integral ( 2 ln(x) dx )

And let's factor out the 2 from the integral.

x [ln(x)]^2 - 2 * Integral ( ln(x) dx )

From here, you would use integration by parts again.

Let u = ln(x). dv = dx.
du = (1/x) dx. v = x

x [ln(x)]^2 - 2[ x ln(x) - Integral ( x(1/x) dx )

Distribute the -2, to get

x [ln(x)]^2 - 2x ln(x) + 2 * Integral ( x(1/x) dx )

Simplify the integral,

x [ln(x)]^2 - 2x ln(x) + 2 * Integral ( 1 dx )

And now it's an easy integral. With no more integrals to solve, don't forget to add a constant.

x [ln(x)]^2 - 2x ln(x) + 2x + C

Factor the common monomial x from the three terms.

x ( [ ln(x)]^2 - 2 ln(x) + 2 ) + C