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

2. Originally Posted by wolfhound
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
Your values for A & B are correct.
But you did not substitute this values in the original equation correctly.

3. Hello miss,
Can you tell me how please?

4. Originally Posted by wolfhound
Hello miss,
Can you tell me how please?
You have the equation:
$\frac{1}{x^2-2x-1}=\frac{A}{x-1+\sqrt{2}}+\frac{B}{x-1-\sqrt{2}}$

And $A=\frac{1}{-2\sqrt{2}} \,\ B=\frac{1}{2\sqrt{2}}$

So when you substitute these values of A&B in the equation, what will you get?

5. Originally Posted by Miss
You have the equation:
$\frac{1}{x^2-2x-1}=\frac{A}{x-1+\sqrt{2}}+\frac{B}{x-1-\sqrt{2}}$

And $A=\frac{1}{-2\sqrt{2}} \,\ B=\frac{1}{2\sqrt{2}}$

So when you substitute these values of A&B in the equation, what will you get?
I dont know what way to arrange them on top of the fractions?
Is it 1/(x-1-Sqrt2) + 1/(x-1+Sqrt2)

6. Originally Posted by wolfhound
I dont know what way to arrange them on top of the fractions?
Is it 1/(x-1-Sqrt2) + 1/(x-1+Sqrt2)
Don't make it complicated
Just replace A by its value and B by its value.
It will be :
$\frac{ \frac{1}{-2\sqrt{2}} }{ x-1+\sqrt{2} } + \frac{ \frac{1}{2\sqrt{2}} }{ x - 1 - \sqrt{2} }$.

To make our life easier, we can write it as:

$\frac{-1}{2\sqrt{2}} \, \frac{1}{x-1+\sqrt{2}} + \frac{1}{2\sqrt{2}} \, \frac{1}{x-1-\sqrt{2}}$

So:

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

Can you integrate this?

7. Originally Posted by Miss
Don't make it complicated
Just replace A by its value and B by its value.
It will be :
$\frac{ \frac{1}{-2\sqrt{2}} }{ x-1+\sqrt{2} } + \frac{ \frac{1}{2\sqrt{2}} }{ x - 1 - \sqrt{2} }$.

To make our life easier, we can write it as:

$\frac{-1}{2\sqrt{2}} \, \frac{1}{x-1+\sqrt{2}} + \frac{1}{2\sqrt{2}} \, \frac{1}{x-1-\sqrt{2}}$

So:

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

Can you integrate this?
Hello miss
Thanks, I can look up how to integrate this,but you could show me if you like

8. Originally Posted by wolfhound
Hello miss
Thanks, I can look up how to integrate this,but you could show me if you like
$\int \left( \frac{-1}{2\sqrt{2}} \, \frac{1}{x-1+\sqrt{2}} + \frac{1}{2\sqrt{2}} \, \frac{1}{x-1-\sqrt{2}} \right) \, dx$

$=\frac{-1}{2\sqrt{2}} \int \frac{dx}{x-1+\sqrt{2}} \, + \frac{1}{2\sqrt{2}} \int \frac{dx}{x-1-\sqrt{2}} \,$

Can you complete it ?

9. Originally Posted by Miss
$\int \left( \frac{-1}{2\sqrt{2}} \, \frac{1}{x-1+\sqrt{2}} + \frac{1}{2\sqrt{2}} \, \frac{1}{x-1-\sqrt{2}} \right) \, dx$

$=\frac{-1}{2\sqrt{2}} \int \frac{dx}{x-1+\sqrt{2}} \, + \frac{1}{2\sqrt{2}} \int \frac{dx}{x-1-\sqrt{2}} \,$

Can you complete it ?
Yes I have it
Thanks

10. You could compare the answer you get to

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

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

which is of the form

$\int{\frac{1}{a^2-u^2}}du=\frac{1}{2a}ln|\frac{a+u}{a-u}|$

$=-\frac{1}{2\sqrt{2}}ln|\frac{\sqrt{2}+x-1}{\sqrt{2}-x+1}|$