Hello Please help me,
Can someone tell me if I have gone wrong in my attached drawing,I have to do integration by partial fractions,also any tips on methods to do partial fraction method would be great, I am not very good at them
Hello Please help me,
Can someone tell me if I have gone wrong in my attached drawing,I have to do integration by partial fractions,also any tips on methods to do partial fraction method would be great, I am not very good at them
Don't make it complicated
Just replace A by its value and B by its value.
It will be :
$\displaystyle \frac{ \frac{1}{-2\sqrt{2}} }{ x-1+\sqrt{2} } + \frac{ \frac{1}{2\sqrt{2}} }{ x - 1 - \sqrt{2} }$.
To make our life easier, we can write it as:
$\displaystyle \frac{-1}{2\sqrt{2}} \, \frac{1}{x-1+\sqrt{2}} + \frac{1}{2\sqrt{2}} \, \frac{1}{x-1-\sqrt{2}}$
So:
$\displaystyle \int \frac{1}{x^2-2x-1} \, dx = \int \left( \frac{-1}{2\sqrt{2}} \, \frac{1}{x-1+\sqrt{2}} + \frac{1}{2\sqrt{2}} \, \frac{1}{x-1-\sqrt{2}} \right) \, dx$.
Can you integrate this?
$\displaystyle \int \left( \frac{-1}{2\sqrt{2}} \, \frac{1}{x-1+\sqrt{2}} + \frac{1}{2\sqrt{2}} \, \frac{1}{x-1-\sqrt{2}} \right) \, dx $
$\displaystyle =\frac{-1}{2\sqrt{2}} \int \frac{dx}{x-1+\sqrt{2}} \, + \frac{1}{2\sqrt{2}} \int \frac{dx}{x-1-\sqrt{2}} \, $
Can you complete it ?
You could compare the answer you get to
$\displaystyle \int{\frac{1}{x^2-2x-1}}dx=\int{\frac{1}{x^2-2x+1-2}}dx=\int{\frac{1}{(x-1)^2-2}}dx$
$\displaystyle =\int{\frac{1}{(x-1)^2-(\sqrt{2})^2}}dx=\int{\frac{-1}{(\sqrt{2})^2-(x-1)^2}}dx$
which is of the form
$\displaystyle \int{\frac{1}{a^2-u^2}}du=\frac{1}{2a}ln|\frac{a+u}{a-u}|$
$\displaystyle =-\frac{1}{2\sqrt{2}}ln|\frac{\sqrt{2}+x-1}{\sqrt{2}-x+1}|$