# Integration

#### Rido12

##### Well-known member
MHB Math Helper

How did WolframAlpha get from step 1 to 2?

I don't get where that extra "two" came from (outside of the integral), and I don't get how there is an extra u factored out.

My initial thought was maybe they mutlipled the outside by two, and inside by 1/2 so that equals 1, but it still doesn't account for the missing 1/sqrt(x) to complete "du".

#### agentmulder

##### Active member
From

$du = \frac{1}{2 \sqrt{x}}dx$

We get $2 \sqrt{x} \ du = dx$

$$2u \ du = dx$$

Now substitute and pull out the constant 2 ...

#### Rido12

##### Well-known member
MHB Math Helper
How did I miss that the first time The ramifications of doing calculus up all night for fun.
Thanks!

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
A simpler approach would be to expand .

#### Rido12

##### Well-known member
MHB Math Helper
A simpler approach would be to expand .
Very true, but I wanted to practice without it Thanks guys!