Integral of 1/1+sqrt(x)

  • #1
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?
 

Answers and Replies

  • #2
Nevermind. I did it now.

It was a simple mistake on my part.

The integral should look like:

[tex]\int_0^1{\frac{2u}{1+u}du}[/tex]

after substitution

and then it should be calculated with integration by parts from there.
 
  • #3
341
0
[tex]\left.{\frac{1}{1+\sqrt{x}}\right|_{x=1} = \left\{\frac{1}{2},\infty\right\}[/tex]
 
  • #4
Gib Z
Homework Helper
3,346
5
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
341
0
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
Ah, yes. That is much easier :) Thanks
 

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