Integrate:
1/(x^(1/2)+x)
I'm not really sure how to "complete the square"
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
Oh. That bit was just illustrating how that was a perfect square. That really wasn't the equation.
Yup.For my problem so far I've got:
1/(z^2+z+(1/2)^2-(1/2)^2)
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
Alternatively
For $\displaystyle \int \frac{1}{x^{1/2}+x} dx$
Taking x^1/2 common factor simplifies things:
$\displaystyle \int \frac{1}{x^{\frac{1}{2}}(1+x^{\frac{1}{2}})}dx$
$\displaystyle u = 1+x^{\frac{1}{2}}$
$\displaystyle 2du = \frac{1}{x^{\frac{1}{2}}}dx$