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

3 Answers
Apr 13, 2015

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#

Apr 13, 2015

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#.

Apr 13, 2015

#=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#