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

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Answer 1 · 2015-04-13T15:14:11

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 · 2015-04-13T15:16:15

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 · 2015-04-13T16:16:41

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

Detail

$int6xlnxdx$

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

$=3x^2lnx-3intxdx$

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

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

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