Question

How do you integrate `int (lnx)^2` by parts?

Answer 1

The answer is `=x((lnx)^2-2lnx+2)+C`

Explanation:

The integration by parts is

`intu'vdx=uv-intuv'dx`

Let `v=(lnx)^2`, `=>`, `v'=(2lnx)/x`

`u'=1`, `=>`, `u=x`

Therefore,

`int(lnx)^2dx=x(lnx)^2-2intlnxdx`

We do theintegration by parts a second time

Let `v=lnx`, `=>`, `v'=1/x`

`u'=1`, `=>`, `u=x`

`intlnxdx=xlnx-intx*1/x*dx`

`=xlnx-x`

Therefore,

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

`=x(lnx)^2-2xlnx+2x+C`

Answer 2

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

Explanation:

The formula for integration by parts states that:

`int u*dv = u*v -int v*du`

In this case we take `u(x) = (lnx)^2` and `v(x) = x`, so that:

`int (lnx)^2dx = x(lnx)^2-int 2xlnx(1/x)dx= x(lnx)^2-2int lnxdx`

We solve this last integral again by parts:

`int lnx = xlnx - int x*(1/x)dx = xlnx -int dx = xlnx -x+C`

Plugging this in the previous result:

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