Question

integral of (ln x)^2

Answer 1

just go int[(lnx)^2]dx = int(2lnx)dx = 2int(lnx)dx

Answer 2

then you just work out the integral of lnx which is fairly straight forward

Answer 3

Thanks!

Answer 4

nope its not that easy

Answer 5

But thats wrong...I think we need to integrate in parts

Answer 6

yes we need by parts

Answer 7

let
\[u=[\ln(x)]^2\\and\\dv=dx ~~which ~~means\\ v=x\]
now use by parts

Answer 8

Thank you!

Answer 9

no problem :)
just use the formula and u'll get the answer

Answer 10

no. you don't need IBP

Answer 11

my apologies

Answer 12

you are correct. use IBP for the integral of lnx

Answer 13

can u do it or need some help

Answer 14

Got it. Thanks guys!

Answer 15

Integral ( [ln(x)]^2 dx )
use integration by parts.
Let u = [ ln(x) ]^2.
dv = dx.
du = 2ln(x) (1/x) dx.

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

And, to make the answer look like yours, let's factor the common monomial x from the three terms.

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