Need help with integration by parts

[Charlesss]

?ln(2x+1) dx
u= ln(2x+1) dv= dx
du= 1/(ln(2x+1)) v= x

xln(2x+1) - ?x/(2x+1) dx

w= x dv=1/(2x+1)
dw=dx v= 1/2 (ln (2x+1))
xln(2x+1) - [ x/2(ln(2x+1) - ?1/2(ln (2x+1)]


..... im lost here the right answer is 1/2(2x+1)ln(2x+1)-x + c

 

[soroban]

Hello, Charlesss!

\(\displaystyle \int\ln(2x+1)\,dx\)

\(\displaystyle u\:=\:\ln(2x+1)\qquad dv\:=\:dx\)

\(\displaystyle du\:=\:\underbrace{\frac{dx}{\ln(2x+1)}}_{no!}\qquad v\:=\: x\)

. . \(\displaystyle \begin{array}{ccccccc}u &=& \ln(2x+1) & & dv &=& dx \\ \\[-3mm] du &=&\frac{2\,dx}{2x+1} & & v &=& x \end{array}\)

\(\displaystyle \text{Then we have: }\;x\cdot\ln(2x+1) - \int \frac{2x}{2x+1}\,dx \;=\;x\cdot\ln(2x+1) - \int\left(1 - \frac{1}{2x+1}\right)dx\)

Got it?