# Thread: what am i doing wrong with this partial fraction integral?

1. ## what am i doing wrong with this partial fraction integral?

(8x^3+7)/((x+1)(2x+1)^3)

I get 8x^3= A(2x+1)^6 + B(x+1)(2x+1)^5 + C(x+1)(2x+1)^4 + D(x+1)(2x+1)^3

i get a=-1 and b,c,d=-1/2

i then integrated each individual fraction and got the wrong answer.. somthing is wrong with my above step. can someone help me to solve it properly and why

any help is very appreciated

2. I get 8x^3= A(2x+1)^6 + B(x+1)(2x+1)^5 + C(x+1)(2x+1)^4 + D(x+1)(2x+1)^3
It may be as simple as noting on the left you should have 8x^3 + 7

If the +7 is missing that will make a difference

3. Originally Posted by twostep08
(8x^3+7)/((x+1)(2x+1)^3)

I get 8x^3= A(2x+1)^6 + B(x+1)(2x+1)^5 + C(x+1)(2x+1)^4 + D(x+1)(2x+1)^3

i get a=-1 and b,c,d=-1/2

i then integrated each individual fraction and got the wrong answer.. somthing is wrong with my above step. can someone help me to solve it properly and why

any help is very appreciated

It is all wrong...or not, above: you wrote $8x^3 + 7$ in the numerator, but then you write only $8x^3$ in the decomposition in partial fractions...and the denominator was $(x+1)(2x+1)^3$, so from where did those powers 6,5,4 come???
Try to write ACCURATELY what you mean.

Tonio

4. yeah i meant to put the 2nd plus 7...i dont know how i missed it-its on my paper. there must be something else wrong

5. the tutor at the library told me to do
A/x+1 + B/2x+1 + C/(2x+1)^2 + D/(2x+1)^3

and then i got the common factor...

its obviously wrong because i got the wrong answer, but if someone can help me with the hard part--ill be able to integrate each seperate fraction (hopefully)

6. See attachment