anonymous
  • anonymous
Calculate the following integral.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
freckles
  • freckles
well we could do partial fractions again
freckles
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

freckles
  • freckles
so it won't be too terrible
freckles
  • 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
freckles
  • freckles
Do you want to try to find A,B,C, and D ?
anonymous
  • anonymous
i got (A+B+C)X^3 + (A-B)x^2 + (A+B-C+D)x +(A-B-D) = x^3−2
anonymous
  • anonymous
SO
anonymous
  • anonymous
(A+B+C)=1 A+B=0 (A+B-C+D)=0 (A-B-D)=-2
Zarkon
  • Zarkon
you are working too hard
anonymous
  • anonymous
:\ I only know how to expand and regroup
freckles
  • freckles
\[A(x^3+x^2+x+1)+B(x^3-x^2+x-1)+C(x^3-x)+D(x^2-1)=x^3-2 \\ x^3(A+B+C)+x^2(A-B+D)+x(A+B-C)+(A-B-D)=x^3-2 \\ A+B+C=1 \\ A-B+D=0 \\ A+B-C=0 \\ A-B-D=-2\]
Zarkon
  • 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
Zarkon
  • 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
Zarkon
  • Zarkon
let x=-1 and you can get B let x=0 you can get d let x=i and you can get C
freckles
  • 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: \[2A+(C+D)=1 \\ 2A-(C+D)=-2 \\ \text{ solving for } A \text{ by adding these two equations together } \\ 4A=-1 \\ A=\frac{-1}{4} \\ \text{ now see the equations that just have } A,B,C \text{ we can solve for } B \text{ and } C \text{ now } \\ A+B+C=1 \\ A+B-C=0 \\ \text{ so } A=\frac{-1}{4} \\ \frac{-1}{4}+B+C=1 \\ \frac{-1}{4}+B-C=0 \text{ adding equations together } \\ \frac{-2}{4}+2B=1 \\ 2B=\frac{3}{2} \\ B=\frac{3}{4} \\ \text{ then find } C \text{ using this system } \\ \text{ then you can finally find } D\]
anonymous
  • anonymous
i got the integral! thank you!

Looking for something else?

Not the answer you are looking for? Search for more explanations.