Question

does anyone know how to find the integral of ln(2x+1)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

\[I=\frac{1}{2}\int\limits_{}{}\ln w dw\]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.

Answer 5

yeah, i used y for the same reason

Answer 6

\[ u=\ln w \rightarrow du = \frac{dw}{w}\]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.

Answer 7

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

Answer 8

most likley