- #1
dragonblood
- 22
- 0
I'm having trouble calculating an integral:
[tex]\int_0^1{\frac{1}{1+\sqrt{x}}dx}[/tex]
I decided to do a substitution:
[tex]u=\sqrt{x}[/tex]
[tex]du=\frac{1}{2\sqrt{x}}dx=\frac{1}{2u}dx[/tex]
thus making the integral look like this:
[tex]\int_0^1{\frac{1}{2u}\cdot\frac{1}{1+u}du}[/tex]
I transformed this integral to:
[tex]\int_0^1{\frac{1}{2u}\cdot\frac{1}{1+u}du}=\int_0^1{\frac{A}{2u}+\frac{B}{1+u}du=\int_0^1{\frac{1}{2u}+\frac{(-1/2)}{1+u}du[/tex]
And tried to solve the integral like this:
[tex]\ln|2u|-\frac{1}{2}\ln|1+u|[/tex]
going from 0 to 1.
However, this means to insert 0 in an ln-expression. Obviously I have done something wrong? Can someone please help?
[tex]\int_0^1{\frac{1}{1+\sqrt{x}}dx}[/tex]
I decided to do a substitution:
[tex]u=\sqrt{x}[/tex]
[tex]du=\frac{1}{2\sqrt{x}}dx=\frac{1}{2u}dx[/tex]
thus making the integral look like this:
[tex]\int_0^1{\frac{1}{2u}\cdot\frac{1}{1+u}du}[/tex]
I transformed this integral to:
[tex]\int_0^1{\frac{1}{2u}\cdot\frac{1}{1+u}du}=\int_0^1{\frac{A}{2u}+\frac{B}{1+u}du=\int_0^1{\frac{1}{2u}+\frac{(-1/2)}{1+u}du[/tex]
And tried to solve the integral like this:
[tex]\ln|2u|-\frac{1}{2}\ln|1+u|[/tex]
going from 0 to 1.
However, this means to insert 0 in an ln-expression. Obviously I have done something wrong? Can someone please help?