## anonymous 5 years ago does anyone know how to find the integral of ln(2x+1)

1. anonymous

Make a substitution of u=(2x+1), find du and use integration by parts.

2. anonymous

so i end up wiht an integral of 1/2ln(u).du, but then what?

3. anonymous

u=2x+1 then du=2dx then dx = du/2, so, letting your integral be I,$I=\frac{1}{2}\int\limits_{}{}\ln u du$

4. anonymous

$I=\frac{1}{2}\int\limits_{}{}\ln w dw$I've just made a switch on the dummy variable since I want to use convention to integrate with IBP, and this convention uses u.

5. anonymous

yeah, i used y for the same reason

6. anonymous

$u=\ln w \rightarrow du = \frac{dw}{w}$and$dv=dw \rightarrow v=w$Then$2I=\ln w .w-\int\limits_{}{}w \frac{dw}{w}=w \ln w - w + c$Since $w=2x+1$$2I=(2x+1)\ln (2x+1) -(2x+1)+c$Divide by 2 to get I.

7. anonymous

yeah, thats what i got, teh answer in the back of my text book gives $1/2(2x+1)\ln (2x+1)-x+C$ but i guess they just expanded the last (2x+1)/2 and included the 1/2 with C

8. anonymous

most likley