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Integration

Please help me with this integral:
(2x^2-x)/((x-2)(x+1))
I have tried partial fractions but this hasn't worked

@Zacken
Original post by loooolo12345
Please help me with this integral:
(2x^2-x)/((x-2)(x+1))
I have tried partial fractions but this hasn't worked


Expand denominator and use long division would be my first guess.
Reply 2
Avatar for Zacken
Zacken
22
Original post by loooolo12345
Please help me with this integral:
(2x^2-x)/((x-2)(x+1))
I have tried partial fractions but this hasn't worked

@Zacken


You need to do long division first because the numerator is of the same degree as the denominator: see this. Is that helpful? If not, come back after you've watched it and show me your working out.
Original post by loooolo12345
Please help me with this integral:
(2x^2-x)/((x-2)(x+1))
I have tried partial fractions but this hasn't worked

@Zacken


its improper, did you use long division first?

Edit: maths forum is 2 quick :frown:

Spoiler

Photo on 07-03-2016 at 19.09 #2.jpg

I have tried doing this
Original post by loooolo12345
Photo on 07-03-2016 at 19.09 #2.jpg

I have tried doing this


Nah, long division first!
(edited 8 years ago)
Reply 6
Avatar for Zacken
Zacken
22
Original post by loooolo12345
Photo on 07-03-2016 at 19.09 #2.jpg

I have tried doing this


You can't do partial fractions unless the degree of the denominator exceed the degree of the numerator. See the video I linked.
Original post by Zacken
You can't do partial fractions unless the degree of the denominator exceed the degree of the numerator. See the video I linked.


ok, thanks, So I have watched the video and now have that the fraction is equal to:

2+((x+4))/((x+1)(x-2))

Is this right?
Reply 8
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Zacken
22
Original post by loooolo12345
ok, thanks, So I have watched the video and now have that the fraction is equal to:

2+((x+4))/((x+1)(x-2))

Is this right?


That's very much correct. Now you can do partial fractions on the second term because the degree of the bottom exceeds the degree of the top.
Original post by Zacken
That's very much correct. Now you can do partial fractions on the second term because the degree of the bottom exceeds the degree of the top.


Ok awesome, I can do the rest now :biggrin: Thank you very much guys
Reply 10
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Zacken
22
Original post by loooolo12345
Ok awesome, I can do the rest now :biggrin: Thank you very much guys


Glad to have helped!
If the degree is the same there should also be constant, so your approach for partial fractions wasn't correct


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Original post by drandy76
If the degree is the same there should also be constant, so your approach for partial fractions wasn't correct


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It was all correct. Have another check!
Original post by Bath_Student
It was all correct. Have another check!


In their initial working?


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Original post by drandy76
In their initial working?


Posted from TSR Mobile


Indeed, I see. The working was redundant though!
Original post by Bath_Student
Indeed, I see. The working was redundant though!


Adding the constant instead of using long division is how I was taught, didn't even realise that was a method of doing it until seeing this thread


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Reply 16
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Zacken
22
Original post by drandy76
Adding the constant instead of using long division is how I was taught, didn't even realise that was a method of doing it until seeing this thread


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They're equivalent. :tongue:
Original post by Zacken
They're equivalent. :tongue:

the second i learned how to equate coefficients i refused to do long division ever again, no ragrets:colone:

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