# Integral of 1/1+sqrt(x)

1. Oct 12, 2009

### 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?

2. Oct 12, 2009

### 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.

3. Oct 12, 2009

### trambolin

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

4. Oct 12, 2009

### 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.

5. Oct 12, 2009

### 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...

6. Oct 12, 2009

### dragonblood

Ah, yes. That is much easier :) Thanks