You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
integration by parts: int ln(2x+1) dx, w/ u=lndx, dv=2x+1
- Thread starter sy211006
- Start date
\(\displaystyle \int ln(2x+1)dx\)
You can not use the choices you have.
Let \(\displaystyle u=ln(2x+1), \;\ dv=dx, \;\ du=\frac{2}{2x+1}dx, \;\ v=x\)
\(\displaystyle xln(2x+1)-2\int\frac{x}{2x+1}dx\)
Now, can you finish by integrating the last one?.
Substitution would be easier, unless you have to use parts.
You can not use the choices you have.
Let \(\displaystyle u=ln(2x+1), \;\ dv=dx, \;\ du=\frac{2}{2x+1}dx, \;\ v=x\)
\(\displaystyle xln(2x+1)-2\int\frac{x}{2x+1}dx\)
Now, can you finish by integrating the last one?.
Substitution would be easier, unless you have to use parts.
D
Deleted member 4993
Guest
galactus said:\(\displaystyle xln(2x+1)-2\int\frac{x}{2x+1}dx\)
\(\displaystyle =\, xln(2x+1)-\int\frac{2x+1 \, - \, 1}{2x+1}dx\)
DUPLICATE POST
viewtopic.php?f=3&t=32795&p=127514#p127514