find or evaluate the integral

**ugkwan** · September 25th 2010, 09:22 AM

Hi

$\displystyle \int \frac{1}{1 + \sqrt{x}}~dx$

I have tried multiplying the numerator with $+\sqrt{x}$ and $-\sqrt{x}$ to factor out the denominator and got stuck
I have tried substituted $1 + \sqrt{x}$ and got stuck
I have tried substituting $\sqrt{x}$ with u and replaced the numerator with $u - \sqrt{x}$ and came up with nothing...

Insights please.

**yeKciM** · September 25th 2010, 09:27 AM

Originally Posted by ugkwan

Hi

$\displystyle \int \frac{1}{1 + \sqrt{x}}~dx$

I have tried multiplying the numerator with $+\sqrt{x}$ and $-\sqrt{x}$ to factor out the denominator and got stuck
I have tried substituted $1 + \sqrt{x}$ and got stuck
I have tried substituting $\sqrt{x}$ with u and replaced the numerator with $u - \sqrt{x}$ and came up with nothing...

Insights please.

use $u = \sqrt {x}$ so $du = \frac {1}{2\sqrt{x}}$

so now your integral will be

$\displaystyle 2\int \frac {u}{u+1} du = \displaystyle 2 \int (1 - \frac {1}{1+u} ) du$

can you continue from here ?

**ugkwan** · September 25th 2010, 09:51 AM

so what is the antiderivative of $\frac {u}{u+1}$ ?

**ugkwan** · September 25th 2010, 10:47 AM

nevermind my previous question. I figured out that by adding 1 and minusing 1 to the numerator, i can split the function, and then proceed with the antiderivative process.

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