- #1
jrmed13
- 14
- 0
Integrating Natural Log Function using "Integration by Parts" Method
The problem says to integrate ln(2x+1)dx
I used u=ln(2x+1); du = 2dx/(2x+1); dv=dx; v=x
So, I integrated it using that (above) 'dictionary' and I got the expression xln(2x+1) - integral of (2x/2x+1)
I could substitute again and say that u=(1/(2x+1)); dv=2xdx, but that process would never end!
And, if I use u=(2x), then various parts of the equation would cancel and I would be left with integral of (ln(2x+1)) = integral of (ln(2x+1))...
I know that the answer should be 0.5(2x+1)ln(2x+1) - x +C, but I can't seem to get it!
Homework Statement
The problem says to integrate ln(2x+1)dx
Homework Equations
I used u=ln(2x+1); du = 2dx/(2x+1); dv=dx; v=x
The Attempt at a Solution
So, I integrated it using that (above) 'dictionary' and I got the expression xln(2x+1) - integral of (2x/2x+1)
I could substitute again and say that u=(1/(2x+1)); dv=2xdx, but that process would never end!
And, if I use u=(2x), then various parts of the equation would cancel and I would be left with integral of (ln(2x+1)) = integral of (ln(2x+1))...
I know that the answer should be 0.5(2x+1)ln(2x+1) - x +C, but I can't seem to get it!