C Charlesss New member Joined Nov 10, 2008 Messages 7 Mar 4, 2009 #1 ?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
?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
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Mar 5, 2009 #2 Hello, Charlesss! \(\displaystyle \int\ln(2x+1)\,dx\) \(\displaystyle u\:=\:\ln(2x+1)\qquad dv\:=\:dx\) \(\displaystyle du\:=\:\underbrace{\frac{dx}{\ln(2x+1)}}_{no!}\qquad v\:=\: x\) Click to expand... . . \(\displaystyle \begin{array}{ccccccc}u &=& \ln(2x+1) & & dv &=& dx \\ \\[-3mm] du &=&\frac{2\,dx}{2x+1} & & v &=& x \end{array}\) \(\displaystyle \text{Then we have: }\;x\cdot\ln(2x+1) - \int \frac{2x}{2x+1}\,dx \;=\;x\cdot\ln(2x+1) - \int\left(1 - \frac{1}{2x+1}\right)dx\) Got it?
Hello, Charlesss! \(\displaystyle \int\ln(2x+1)\,dx\) \(\displaystyle u\:=\:\ln(2x+1)\qquad dv\:=\:dx\) \(\displaystyle du\:=\:\underbrace{\frac{dx}{\ln(2x+1)}}_{no!}\qquad v\:=\: x\) Click to expand... . . \(\displaystyle \begin{array}{ccccccc}u &=& \ln(2x+1) & & dv &=& dx \\ \\[-3mm] du &=&\frac{2\,dx}{2x+1} & & v &=& x \end{array}\) \(\displaystyle \text{Then we have: }\;x\cdot\ln(2x+1) - \int \frac{2x}{2x+1}\,dx \;=\;x\cdot\ln(2x+1) - \int\left(1 - \frac{1}{2x+1}\right)dx\) Got it?
