Calculate the following integral.

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Calculate the following integral.

Mathematics
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

1 Attachment
well we could do partial fractions again
\[x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)\] there is only 3 factors to worry about on the denominator

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

so it won't be too terrible
\[\frac{x^3-2}{x^4-1}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1} \\ x^3-2=A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)(x+1)\] so multiply out and reorder then compare both sides
Do you want to try to find A,B,C, and D ?
i got (A+B+C)X^3 + (A-B)x^2 + (A+B-C+D)x +(A-B-D) = x^3−2
SO
(A+B+C)=1 A+B=0 (A+B-C+D)=0 (A-B-D)=-2
you are working too hard
:\ I only know how to expand and regroup
\[A(x^3+x^2+x+1)+B(x^3-x^2+x-1)+C(x^3-x)+D(x^2-1)=x^3-2 \\ x^3(A+B+C)+x^2(A-B+D)+x(A+B-C)+(A-B-D)=x^3-2 \\ A+B+C=1 \\ A-B+D=0 \\ A+B-C=0 \\ A-B-D=-2\]
\[\frac{x^3-2}{x^4-1}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1} \\ x^3-2=A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)(x+1)\] let x=1
\[1^3-2=A(1+1)(1^2+1)+B(1-1)(1^2+1)+(C1+D)(1-1)(1+1)\] \[1-2=A(1+1)(1+1)\] A=-1/4
let x=-1 and you can get B let x=0 you can get d let x=i and you can get C
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: \[2A+(C+D)=1 \\ 2A-(C+D)=-2 \\ \text{ solving for } A \text{ by adding these two equations together } \\ 4A=-1 \\ A=\frac{-1}{4} \\ \text{ now see the equations that just have } A,B,C \text{ we can solve for } B \text{ and } C \text{ now } \\ A+B+C=1 \\ A+B-C=0 \\ \text{ so } A=\frac{-1}{4} \\ \frac{-1}{4}+B+C=1 \\ \frac{-1}{4}+B-C=0 \text{ adding equations together } \\ \frac{-2}{4}+2B=1 \\ 2B=\frac{3}{2} \\ B=\frac{3}{4} \\ \text{ then find } C \text{ using this system } \\ \text{ then you can finally find } D\]
i got the integral! thank you!

Not the answer you are looking for?

Search for more explanations.

Ask your own question