Integration Please help,

**wolfhound** · March 4th 2010, 10:14 AM

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

**Miss** · March 4th 2010, 10:22 AM

Originally Posted by wolfhound

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

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

**wolfhound** · March 4th 2010, 10:28 AM

Hello miss,
Can you tell me how please?

**Miss** · March 4th 2010, 10:40 AM

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?

**wolfhound** · March 4th 2010, 10:42 AM

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?
Help please! Thanks
Is it 1/(x-1-Sqrt2) + 1/(x-1+Sqrt2)

**Miss** · March 4th 2010, 10:53 AM

Originally Posted by wolfhound

I dont know what way to arrange them on top of the fractions?
Help please! Thanks
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?

**wolfhound** · March 4th 2010, 11:05 AM

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

Thanks for your help!

**Miss** · March 4th 2010, 11:29 AM

Originally Posted by wolfhound

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

Thanks for your help!

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

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

Can you complete it ?

**wolfhound** · March 4th 2010, 12:48 PM

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

**Archie Meade** · March 4th 2010, 01:46 PM

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

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

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

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