Struggling with Calculating an Integral: Can Someone Help?

[dragonblood]

I'm having trouble calculating an integral:

\int_0^1{\frac{1}{1+\sqrt{x}}dx}

I decided to do a substitution:

u=\sqrt{x}
du=\frac{1}{2\sqrt{x}}dx=\frac{1}{2u}dx
thus making the integral look like this:
\int_0^1{\frac{1}{2u}\cdot\frac{1}{1+u}du}
I transformed this integral to:
\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
And tried to solve the integral like this:
\ln|2u|-\frac{1}{2}\ln|1+u|
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?

 

[dragonblood]

Nevermind. I did it now.

It was a simple mistake on my part.

The integral should look like:

\int_0^1{\frac{2u}{1+u}du}

after substitution

and then it should be calculated with integration by parts from there.

 

[trambolin]

\left.{\frac{1}{1+\sqrt{x}}\right|_{x=1} = \left\{\frac{1}{2},\infty\right\}

 

[Gib Z]

I'm not entirely sure what that post means trambolin, care to explain?

Dragonblood - The solution comes out must quicker than integration by parts if you add and subtract 2 into the numerator.

 

[trambolin]

I meant the -1 is causing the integrand blow up at x=1 for the cause of ln 0. But nevermind it is just the notation stuff...

 

[dragonblood]

Ah, yes. That is much easier :) Thanks