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About LinkBacksHello 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
You have the equation:Originally Posted by wolfhound
$\displaystyle \frac{1}{x^2-2x-1}=\frac{A}{x-1+\sqrt{2}}+\frac{B}{x-1-\sqrt{2}}$
And $\displaystyle A=\frac{1}{-2\sqrt{2}} \,\ B=\frac{1}{2\sqrt{2}}$
So when you substitute these values of A&B in the equation, what will you get?
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?
Hello missDon'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?
Thanks, I can look up how to integrate this,but you could show me if you like
Thanks for your help!
$\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 $Hello miss
Thanks, I can look up how to integrate this,but you could show me if you like
Thanks for your help!
$\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
$\displaystyle \int{\frac{1}{a^2-u^2}}du=\frac{1}{2a}ln|\frac{a+u}{a-u}|$
$\displaystyle =-\frac{1}{2\sqrt{2}}ln|\frac{\sqrt{2}+x-1}{\sqrt{2}-x+1}|$
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