# Thread: Need help with Integration and completing the square

1. ## Need help with Integration and completing the square

Integrate:

1/(x^(1/2)+x)

I'm not really sure how to "complete the square"

2. Originally Posted by plooms
Integrate:

1/(x^(1/2)+x)

I'm not really sure how to "complete the square"
Let $\displaystyle z=x^{\frac{1}{2}}$

So we see that we have $\displaystyle \frac{1}{(x^{\frac{1}{2}})^2+x^{\frac{1}{2}}}=\fra c{1}{z^2+z}$

Can you complete the square now?

--Chris

3. Woah, thanks for that first step. I definitely wouldn't have looked at it that way.

My main problem is I never learned how to "complete the square." I don't even know what that means!

4. Originally Posted by plooms
Woah, thanks for that first step. I definitely wouldn't have looked at it that way.

My main problem is I never learned how to "complete the square." I don't even know what that means!
Let's say you had a quadratic equation of the form $\displaystyle ax^2+bx=0$

To use the complete the square technique, the coefficient of $\displaystyle x^2$ must be 1.

So we can rewrite the equation $\displaystyle x^2+\frac{b}{a}x=0$

Now this is where we apply this technique.

You need to introduce this new term into the equation.

Take the coefficient of the x term, divide by 2, then square that quantity:

$\displaystyle \left(\frac{b}{2a}\right)^2$

In essence, we now add "zero" to the equation:

$\displaystyle x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2=0$.

$\displaystyle x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2$ becomes a perfect square:

$\displaystyle x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2=\left (x+\frac{b}{2a}\right)^2$

So our equation now takes on the general form $\displaystyle \left(x+\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2=0$.

We have "completed the square".

Does this make sense?

--Chris

5. Sort of...

I got lost on the last step before you generalized. Where did you get the x on the RHS?

For my problem so far I've got:

1/(z^2+z+(1/2)^2-(1/2)^2)

6. Originally Posted by plooms
Sort of...

I got lost on the last step before you generalized. Where did you get the x on the RHS?
Oh. That bit was just illustrating how that was a perfect square. That really wasn't the equation.

For my problem so far I've got:

1/(z^2+z+(1/2)^2-(1/2)^2)
Yup.

So that means that we should get $\displaystyle \frac{1}{\left(z+\frac{1}{2}\right)^2-\frac{1}{4}}=\frac{4}{4\left(x+\frac{1}{2}\right)^ 2-1}=\frac{4}{(2x+1)^2-1}$

--Chris

7. Alternatively

For $\displaystyle \int \frac{1}{x^{1/2}+x} dx$

Taking x^1/2 common factor simplifies things:

$\displaystyle \int \frac{1}{x^{\frac{1}{2}}(1+x^{\frac{1}{2}})}dx$

$\displaystyle u = 1+x^{\frac{1}{2}}$
$\displaystyle 2du = \frac{1}{x^{\frac{1}{2}}}dx$

8. alright thanks, i think i've got it from there