integration by parts: int ln(2x+1) dx, w/ u=lndx, dv=2x+1

[sy211006]

i keep going in circles could someone please help me

prob
integral ln(2x+1) dx

i use u=ln dx and dv=2x+1 when i integrate it seems that i just keep going in circles

 

[galactus]

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

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.

 

[sy211006]

thank you
now i will try the rest on my own
i have to use intergration by parts
might have more questions later
wish me luck

 

[sy211006]

ok i am worse than i thought...
please show me the next step

 

[Deleted member 4993]

$\displaystyle xln(2x+1)-2\int\frac{x}{2x+1}dx$

$\displaystyle =\, xln(2x+1)-\int\frac{2x+1 \, - \, 1}{2x+1}dx$

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