Hi all!
I am working on this Integration by partial fraction question and I used the -b formula to factor the bottom line but I'm not sure where to go from here. I would really appreciate any help!
$\displaystyle \int \frac{1}{x^2-2x-1}dx$
Hi all!
I am working on this Integration by partial fraction question and I used the -b formula to factor the bottom line but I'm not sure where to go from here. I would really appreciate any help!
$\displaystyle \int \frac{1}{x^2-2x-1}dx$
Hi NFZ17,
so you have $\displaystyle x^2-2x-1=(x-1-\sqrt{2})(x-1+\sqrt{2})$
then
$\displaystyle \frac{1}{x^2-2x-1}=\frac{A}{x-1-\sqrt{2}}+\frac{B}{x-1+\sqrt{2}}$
$\displaystyle 1=A(x-1+\sqrt{2})+B(x-1-\sqrt{2})$
Finish up by placing $\displaystyle x=1+\sqrt{2}$ to find A, and $\displaystyle x=1-\sqrt{2}$ to find B
Alternatively:
$\displaystyle \int{\frac{1}{x^2 - 2x - 1}\,dx} = \int{\frac{1}{x^2 - 2x + (-1)^2 - (-1)^2 - 1}\,dx}$
$\displaystyle = \int{\frac{1}{(x - 1)^2 - 2}\,dx}$.
Now you can make the substitution $\displaystyle x - 1 = \sqrt{2}\cosh{t}$ so that $\displaystyle dx = \sqrt{2}\sinh{t}\,dt$.
Note that $\displaystyle t = \textrm{arcosh}\,{\left(\frac{\sqrt{2}(x - 1)}{2}\right)}$.
The integral becomes:
$\displaystyle \int{\frac{1}{(\sqrt{2}\cosh{t})^2 - 2}\,\sqrt{2}\sinh{t}\,dt}$
$\displaystyle = \int{\frac{\sqrt{2}\sinh{t}}{2\cosh^2{t} - 2}\,dt}$
$\displaystyle = \int{\frac{\sqrt{2}\sinh{t}}{2(\cosh^2{t} - 1)}\,dt}$
$\displaystyle = \int{\frac{\sqrt{2}\sinh{t}}{2\sinh^2{t}}\,dt}$
$\displaystyle = \int{\frac{\sqrt{2}}{2}\,dt}$
$\displaystyle = \frac{\sqrt{2}}{2}t + C$
$\displaystyle = \frac{\sqrt{2}}{2}\textrm{arcosh}\,{\left(\frac{\s qrt{2}(x - 1)}{2}\right)} + C$.