# Thread: Completing the square and integration

1. ## Completing the square and integration

$\displaystyle \int \frac{5}{x^2+x-1} dx$

I know that $\displaystyle \frac{\sqrt{5}}{2} sec\theta = x + 1/2$

but its so hard for me to get a clear answer.

Can someone spare some time for me and show me how this is done?

Thank You so much

2. $\displaystyle x^2+x-1= \left( x+\frac{1}{2}\right)^2-\frac{5}{4}$

now make the substitution.

3. Originally Posted by pickslides
$\displaystyle x^2+x-1= \left( x+\frac{1}{2}\right)^2-\frac{5}{4}$

now make the substitution.
i know but what im getting is $\displaystyle 2\sqrt{5}\int csc\theta d\theta$

i dont know what to do from there to get a clear and reduced answered

4. multiply by $\displaystyle 1 = \frac{\csc \theta + \cot \theta}{\csc \theta + \cot \theta}$

then you have $\displaystyle 2\sqrt{5}\int \frac{\csc^2 \theta + \csc \theta \cot \theta}{\csc \theta + \cot \theta} d \theta$

which is of the form $\displaystyle \int \frac {du}{u}$.

5. Originally Posted by icemanfan
multiply by $\displaystyle 1 = \frac{\csc \theta + \cot \theta}{\csc \theta + \cot \theta}$

then you have $\displaystyle 2\sqrt{5}\int \frac{\csc^2 \theta + \csc \theta \cot \theta}{\csc \theta + \cot \theta} d \theta$

which is of the form $\displaystyle \int \frac {du}{u}$.
thank you but i know the integral of that but what I'm looking for the answer and how gotten to it.

$\displaystyle -\ln \mid \csc\theta + \cot\theta \mid + c$