# Integration by partial fractions

• March 23rd 2010, 04:42 AM
NFZ17
Integration by partial fractions
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!

$\int \frac{1}{x^2-2x-1}dx$
• March 23rd 2010, 04:56 AM
Quote:

Originally Posted by NFZ17
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!

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

Hi NFZ17,

so you have $x^2-2x-1=(x-1-\sqrt{2})(x-1+\sqrt{2})$

then

$\frac{1}{x^2-2x-1}=\frac{A}{x-1-\sqrt{2}}+\frac{B}{x-1+\sqrt{2}}$

$1=A(x-1+\sqrt{2})+B(x-1-\sqrt{2})$

Finish up by placing $x=1+\sqrt{2}$ to find A, and $x=1-\sqrt{2}$ to find B
• March 23rd 2010, 05:04 AM
Prove It
Quote:

Originally Posted by NFZ17
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!

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

Alternatively:

$\int{\frac{1}{x^2 - 2x - 1}\,dx} = \int{\frac{1}{x^2 - 2x + (-1)^2 - (-1)^2 - 1}\,dx}$

$= \int{\frac{1}{(x - 1)^2 - 2}\,dx}$.

Now you can make the substitution $x - 1 = \sqrt{2}\cosh{t}$ so that $dx = \sqrt{2}\sinh{t}\,dt$.

Note that $t = \textrm{arcosh}\,{\left(\frac{\sqrt{2}(x - 1)}{2}\right)}$.

The integral becomes:

$\int{\frac{1}{(\sqrt{2}\cosh{t})^2 - 2}\,\sqrt{2}\sinh{t}\,dt}$

$= \int{\frac{\sqrt{2}\sinh{t}}{2\cosh^2{t} - 2}\,dt}$

$= \int{\frac{\sqrt{2}\sinh{t}}{2(\cosh^2{t} - 1)}\,dt}$

$= \int{\frac{\sqrt{2}\sinh{t}}{2\sinh^2{t}}\,dt}$

$= \int{\frac{\sqrt{2}}{2}\,dt}$

$= \frac{\sqrt{2}}{2}t + C$

$= \frac{\sqrt{2}}{2}\textrm{arcosh}\,{\left(\frac{\s qrt{2}(x - 1)}{2}\right)} + C$.
• March 23rd 2010, 06:18 AM
NFZ17
Archie, I used your method and came up with...

$\frac{1}{2\sqrt2} ln[x-1+\sqrt2]-\frac{1}{2\sqrt2} ln[x-1-\sqrt2]+c$

Does this look right? I'm a disaster when it comes to surds!
• March 23rd 2010, 06:28 AM
Hi NFZ17,

double check your values for A and B to see if you've got the two
• March 23rd 2010, 06:47 AM
NFZ17
Quote:

Hi NFZ17,

so you have $x^2-2x-1=(x-1-\sqrt{2})(x-1+\sqrt{2})$

then

$\frac{1}{x^2-2x-1}=\frac{A}{x-1-\sqrt{2}}+\frac{B}{x-1+\sqrt{2}}$

$1=A(x-1+\sqrt{2})+B(x-1-\sqrt{2})$

Finish up by placing $x=1+\sqrt{2}$ to find A, and $x=1-\sqrt{2}$ to find B

If I set $x=1+\sqrt{2}$

I get, $A=0$ and, $B=\frac{1}{2\sqrt2}$

and then,

$A=\frac{-1}{2\sqrt2}$

is this where I'm going wrong?
• March 23rd 2010, 06:54 AM
Quote:

Originally Posted by NFZ17
If I set $x=1+\sqrt{2}$

I get, $A=0$ and, $B=\frac{1}{2\sqrt2}$

and then,

$A=\frac{-1}{2\sqrt2}$

is this where I'm going wrong?

If you set $x=1+\sqrt{2}$, you get

$1=A(1+\sqrt{2}-1+\sqrt{2})+B(1+\sqrt{2}-1-\sqrt{2})=A(2\sqrt{2})$

Then $A=\frac{1}{2\sqrt{2}}$

You get A through multiplying B by 0.
Similarly, find B by cancelling A.
• March 23rd 2010, 07:15 AM
NFZ17
Thank Archie!

$\frac{1}{2\sqrt2} ln[x-1-\sqrt2]-\frac{1}{2\sqrt2} ln[x-1+\sqrt2]+c$

I hope this is right!