- #1
Saladsamurai
- 3,020
- 7
So I am using int by parts to evaluate [tex]\int_{-1}^1\ln(x+2)dx[/tex]
I am doing something incorrectly with the indefinite int part of this problem, but I can'y find my error, can you?
[tex]\int udv=uv-\int vdu[/tex]
[tex]u=ln(x+2)[/tex]
[tex]\Rightarrow du=\frac{dx}{x+2}[/tex]
[tex]dv=dx[/tex]
[tex]\Rightarrow v=x[/tex]
So I get:
[tex]x\ln(x+2)-\int\frac{x}{x+2}dx[/tex]
Working on only the last part ^^^by long division I get:
[tex]\int\frac{x}{x+2}dx=\int(1-\frac{2}{x+2})dx[/tex]
[tex]=x-2\ln(x+2)[/tex]
and recombining with the top:
[tex]x\ln(x+2)-x+2\ln(x+2)[/tex]
But when I evaluate this on my calculator, I get the incorrect answer. Any ideas?
Casey
I am doing something incorrectly with the indefinite int part of this problem, but I can'y find my error, can you?
[tex]\int udv=uv-\int vdu[/tex]
[tex]u=ln(x+2)[/tex]
[tex]\Rightarrow du=\frac{dx}{x+2}[/tex]
[tex]dv=dx[/tex]
[tex]\Rightarrow v=x[/tex]
So I get:
[tex]x\ln(x+2)-\int\frac{x}{x+2}dx[/tex]
Working on only the last part ^^^by long division I get:
[tex]\int\frac{x}{x+2}dx=\int(1-\frac{2}{x+2})dx[/tex]
[tex]=x-2\ln(x+2)[/tex]
and recombining with the top:
[tex]x\ln(x+2)-x+2\ln(x+2)[/tex]
But when I evaluate this on my calculator, I get the incorrect answer. Any ideas?
Casey
Last edited: