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# Partial fractions and integration question watch

1. I just can't get the integration part of the question right:

a)express 2x+1/(2x+3)^2 in the form A/2x+3 + B/(2x+3)^2

I got 1/2x+3 - 2/(2x+3)^2

b) hence show that (the integral between 2 and -1) of 2x+1/(2x+3)^2 dx = 1/2ln7 - 6/7

I know that you integrate the partial fractions but i think i've gone wrong somewhere after that
2. (Original post by mica-lwe)
I just can't get the integration part of the question right:

a)express 2x+1/(2x+3)^2 in the form A/2x+3 + B/(2x+3)^2

I got 1/2x+3 - 2/(2x+3)^2

b) hence show that (the integral between 2 and -1) of 2x+1/(2x+3)^2 dx = 1/2ln7 - 6/7

I know that you integrate the partial fractions but i think i've gone wrong somewhere after that
3. (Original post by TeeEm)
^^^there they are
4. (Original post by mica-lwe)
^^^there they are
2nd fraction does not integrate to a log
5. (Original post by TeeEm)
2nd fraction does not integrate to a log
how do you integrate it then?
6. (Original post by mica-lwe)
how do you integrate it then?
write it to the minus 2 and integrate by inspection(recognition)

or use substitution
7. (Original post by TeeEm)
write it to the minus 2 and integrate by inspection(recognition)

or use substitution
to integrate a bracket do you raise the power by one, then divide the bracket by the new power x the differential of the bracket?
8. (Original post by mica-lwe)
to integrate a bracket do you raise the power by one, then divide the bracket by the new power x the differential of the bracket?
yes

(so long as the differential of the bracket is a constant, which is the case here)
9. (Original post by TeeEm)
yes

(so long as the differential of the bracket is a constant, which is the case here)
I almost have the right answer, I have 2ln7 -6/7 but I need 1/2ln7 -6/7, don't know why it's 1/2 instead of 2?
10. (Original post by mica-lwe)
I almost have the right answer, I have 2ln7 -6/7 but I need 1/2ln7 -6/7, don't know why it's 1/2 instead of 2?
after your partial fractions you have put a 2 in front of the integral

when you integrated the log, you have not divided by 2

hopefully when you sort them it will work
11. (Original post by TeeEm)
after your partial fractions you have put a 2 in front of the integral

when you integrated the log, you have not divided by 2

hopefully when you sort them it will work
Thankyou!
12. (Original post by mica-lwe)
Thankyou!
you are welcome

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Updated: December 4, 2014
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