Question

How do you integrate `6x ln x dx`?

Answer 1

Integrate by parts.

`int 6xln x dx`

Let `u = lnx` so `du = 1/x dx` and

let `dv = 6x dx` to get `v=3x^2`

`int udv = uv - int vdu`

`int 6xln x dx = 3x^2 lnx - int 3x^2 1/x dx`

`color(white)"ssssssssssss"` `= 3x^2 lnx - int 3x dx`

`color(white)"ssssssssssss"` `= 3x^2 lnx - 3/2 x^2 +C`

Answer 2

The answer is: `3x^2lnx-3x^2/2+c`.

This integral has to be done with the theorem of the integration by parts, that says:

`intf(x)g'(x)dx=f(x)g(x)-intg(x)f'(x)dx`

We can assume that `f(x)=lnx` and `g'(x)dx=6xdx`.

`g(x)=6x^2/2=3x^2`

`f'(x)dx=1/xdx`.

So:

`int6xlnxdx=3x^2lnx-int3x^2*1/xdx=3x^2lnx-3intxdx=`

`=3x^2lnx-3x^2/2+c`.

Answer 3

`=3x^2lnx-1/4x^2+C`

Detail

`int6xlnxdx`

`=lnx(6(x)^2)/(2)-int(1)/(x).(6x^2)/(2)dx`

`=3x^2lnx-3intxdx`

`=3x^2lnx-3(x^2)/2+C`

`=3x^2lnx-3/2x^2+C`