How do you integrate ln(2x)ln(2x)?

1 Answer
Nov 24, 2016

x(ln(2x)-1)+Cx(ln(2x)1)+C

Explanation:

I=intln(2x)dxI=ln(2x)dx

We should use integration by parts in the absence of all other possible integration strategies. Integration by parts takes the form intudv=uv-intvduudv=uvvdu. For intln(2x)dxln(2x)dx, let:

{(u=ln(2x),=>,du=2/(2x)=1/x),(dv=dx,=>,v=x):}

Thus:

I=uv-intvdu=xln(2x)-intx1/xdx

I=xln(2x)-intdx

I=xln(2x)-x

I=x(ln(2x)-1)+C