- #1

- 22

- 0

[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?