$\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
$\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
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 $\displaystyle \int \frac {du}{u}$.