## anonymous one year ago Calculate the following integral.

1. anonymous

2. freckles

well we could do partial fractions again

3. freckles

$x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)$ there is only 3 factors to worry about on the denominator

4. freckles

so it won't be too terrible

5. freckles

$\frac{x^3-2}{x^4-1}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1} \\ x^3-2=A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)(x+1)$ so multiply out and reorder then compare both sides

6. freckles

Do you want to try to find A,B,C, and D ?

7. anonymous

i got (A+B+C)X^3 + (A-B)x^2 + (A+B-C+D)x +(A-B-D) = x^3−2

8. anonymous

SO

9. anonymous

(A+B+C)=1 A+B=0 (A+B-C+D)=0 (A-B-D)=-2

10. Zarkon

you are working too hard

11. anonymous

:\ I only know how to expand and regroup

12. freckles

$A(x^3+x^2+x+1)+B(x^3-x^2+x-1)+C(x^3-x)+D(x^2-1)=x^3-2 \\ x^3(A+B+C)+x^2(A-B+D)+x(A+B-C)+(A-B-D)=x^3-2 \\ A+B+C=1 \\ A-B+D=0 \\ A+B-C=0 \\ A-B-D=-2$

13. Zarkon

$\frac{x^3-2}{x^4-1}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1} \\ x^3-2=A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)(x+1)$ let x=1

14. Zarkon

$1^3-2=A(1+1)(1^2+1)+B(1-1)(1^2+1)+(C1+D)(1-1)(1+1)$ $1-2=A(1+1)(1+1)$ A=-1/4

15. Zarkon

let x=-1 and you can get B let x=0 you can get d let x=i and you can get C

16. freckles

if you were having trouble solving the system, this may help I would add the first two equations together and the second two equations together: giving you: $2A+(C+D)=1 \\ 2A-(C+D)=-2 \\ \text{ solving for } A \text{ by adding these two equations together } \\ 4A=-1 \\ A=\frac{-1}{4} \\ \text{ now see the equations that just have } A,B,C \text{ we can solve for } B \text{ and } C \text{ now } \\ A+B+C=1 \\ A+B-C=0 \\ \text{ so } A=\frac{-1}{4} \\ \frac{-1}{4}+B+C=1 \\ \frac{-1}{4}+B-C=0 \text{ adding equations together } \\ \frac{-2}{4}+2B=1 \\ 2B=\frac{3}{2} \\ B=\frac{3}{4} \\ \text{ then find } C \text{ using this system } \\ \text{ then you can finally find } D$

17. anonymous

i got the integral! thank you!