Question

Calculate the following integral.

Answer 1

well we could do partial fractions again

Answer 2

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

Answer 3

so it won't be too terrible

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

SO

Answer 8

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

Answer 9

you are working too hard

Answer 10

:\ I only know how to expand and regroup

Answer 11

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

Answer 12

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

let x=1

Answer 13

\[1^3-2=A(1+1)(1^2+1)+B(1-1)(1^2+1)+(C1+D)(1-1)(1+1)\]
\[1-2=A(1+1)(1+1)\]
A=-1/4

Answer 14

let x=-1 and you can get B

let x=0 you can get d

let x=i and you can get C

Answer 15

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:
\[2A+(C+D)=1 \\ 2A-(C+D)=-2 \\ \text{ solving for } A \text{ by adding these two equations together } \\ 4A=-1 \\ A=\frac{-1}{4} \\ \text{ now see the equations that just have } A,B,C \text{ we can solve for } B \text{ and } C \text{ now } \\ A+B+C=1 \\ A+B-C=0 \\ \text{ so } A=\frac{-1}{4} \\ \frac{-1}{4}+B+C=1 \\ \frac{-1}{4}+B-C=0 \text{ adding equations together } \\ \frac{-2}{4}+2B=1 \\ 2B=\frac{3}{2} \\ B=\frac{3}{4} \\ \text{ then find } C \text{ using this system } \\ \text{ then you can finally find } D\]

Answer 16

i got the integral! thank you!