You do this with integration by parts
f(x)=lnx g'(x)=3
f'(x)=1/x g(x)=3x
$$\\\boxed{\int\;f(x)g'(x)dx=f(x)g(x)-\int\;g(x)f'(x)dx}\\\\\\\\
\int\;((lnx)*3)dx\\\\
=(lnx)(3x)-\int\;(3x*\frac{1}{x})dx\\\\
=(3xlnx)-\int\;(3)dx\\\\
=(3xlnx)-3x+c\\\\
=3x(ln(x)-1)+c$$
I think that my working is right but you should check it anyway :)
You do this with integration by parts
f(x)=lnx g'(x)=3
f'(x)=1/x g(x)=3x
$$\\\boxed{\int\;f(x)g'(x)dx=f(x)g(x)-\int\;g(x)f'(x)dx}\\\\\\\\
\int\;((lnx)*3)dx\\\\
=(lnx)(3x)-\int\;(3x*\frac{1}{x})dx\\\\
=(3xlnx)-\int\;(3)dx\\\\
=(3xlnx)-3x+c\\\\
=3x(ln(x)-1)+c$$
I think that my working is right but you should check it anyway :)