Integral of (1/x)(lnx)^2

[st3dent]

I don't know what i'm doing wrong when taking this simple integral.

The integral is:

\int (1/x)(lnx)^2 dx

\int (2/x)(lnx) dx

2 \int (lnx/x) dx

Let u = lnx
du/dx = 1/x
dx = xdu

2 \int (u/x) xdu

2 \int (u) du

2 \int (u^2)/2 + C

(2(lnx)^2)/2 + C

(lnx)^2 + C

The answer is obviously wrong....how do i solve this properly?

[cookiemonster]

Try the substitution u = lnx.

cookiemonster

[HallsofIvy]

I don't know what i'm doing wrong when taking this simple integral.

The integral is:

\int (1/x)(lnx)^2 dx

\int (2/x)(lnx) dx

How did you make that step? ln(x2)= 2ln(x)
but this is (ln(x))2.

Just go ahead and make the substitution u= ln(x) right at the start.

[st3dent]

Thanks..seems to work now. I got it. Thanks for all your help.