anonymous
  • anonymous
Compute the integral.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
1 Attachment
tacotime
  • tacotime
You can change the exponents to negatives to move the denominator to the numerator.
freckles
  • freckles
partial fractions should work

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freckles
  • freckles
\[\frac{1}{x^2(x^2+4)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+4}\]
anonymous
  • anonymous
Yeah i got that , having difficulties finding A,B,C, and D
freckles
  • freckles
which method are you used to: heaviside or comparing and solving a system of linear equations?
freckles
  • freckles
\[Ax(x^2+4)+B(x^2+4)+(Cx+D)(x^2)=1 \]
anonymous
  • anonymous
both were taught i suppose. i dont mind expanding everything
freckles
  • freckles
for example heaviside is pluggin in numbers to help solve like replace x with 0 and we get B(0+4)=1 so B=1/4 then you can plug in 2i for x \[(C \cdot 2i+D)(4i^2)=1 \\ (C \cdot 2i+D)(-4)=1 \\ C \cdot 2i+D=\frac{-1}{4} \] plug in -2i for x \[(C \cdot (-2i)+D)(4i^2)=1 \\ -2i C+D=\frac{-1}{4} \\ 2iC+D=\frac{-1}{4} \text{ which was given by the other x value we plugged \in } \\ \text{ now add } 2D=\frac{-2}{4}=\frac{-1}{2} \\ \text{ so } D=\frac{-1}{4} \\ \text{ so now } C \\ -2i C-\frac{1}{4}=\frac{-1}{4} \\ -2i C=0 \\ \text{ so we have } C=0\]
freckles
  • freckles
but you probably don't like imaginary numbers... we could just do the whole expanding thing then comparing then solving the system thing we ended up solving a system for the heaviside anyways
anonymous
  • anonymous
it's definately not the heaviside lol
freckles
  • freckles
I only plugged in -2i and 2i because it made evaluating x^2+4 simple
freckles
  • freckles
since (2i)^2+4 is 0 and (-2i)^2+4 is also 0
freckles
  • freckles
ok you expand and I will expand and regroup terms and you can compare my equation to your equation
freckles
  • freckles
\[Ax(x^2+4)+B(x^2+4)+(Cx+D)(x^2)=1 \\ A(x^3+4x)+B(x^2+4)+Cx^3+Dx^2 =1 \\ x^3(A+C)+x^2(B+D)+x(4A)+(4B)=1\]
anonymous
  • anonymous
|dw:1444155517948:dw|
freckles
  • freckles
great we have no x^3 's on the other side so you should see A+C=0 we have no x^2 's on the other side so you should see B+D=0 we have no x 's on the other side so you should see 4A=0 we do have constants on both sides so we need 4B=1
freckles
  • freckles
I would solve the last two equations first because they only involve one constant variable each
freckles
  • freckles
\[A=0 , B=\frac{1}{4}\]
freckles
  • freckles
and then go to the top two equations and used these values to solve for C and D
anonymous
  • anonymous
hold on, i always get stuck on this part. you get A+C=0 .B+D=0. 4A=0. 4B=1 by comparing the variable son each side?
freckles
  • freckles
yeah yeah
freckles
  • freckles
\[Ax(x^2+4)+B(x^2+4)+(Cx+D)(x^2)=1 \\ A(x^3+4x)+B(x^2+4)+Cx^3+Dx^2 =1 \\ x^3(A+C)+x^2(B+D)+x(4A)+(4B)=1 \text{ same as } \\ x^3(A+C)+x^2(B+D)+x(4A)+(4B)=x^3(0)+x^2(0)+x(0)+1\]
freckles
  • freckles
we want both sides to be the same thing this can only happen if A+C and B+D and 4A are zero and 4B is one.
anonymous
  • anonymous
A=0 B=1/4 C=0 D=-1/4
freckles
  • freckles
sounds awesome so you have: \[\int\limits \frac{1}{4} \frac{1}{x^2} dx-\int\limits \frac{1}{4} \cdot \frac{1}{x^2+4} dx\]
freckles
  • freckles
\[\frac{1}{4} \int\limits x^{-2} dx-\frac{1}{4} \int\limits \frac{1}{x^2+4} dx \\ \frac{1}{4} \int\limits x^{-2} dx - \frac{1}{4 } \int\limits \frac{\frac{1}{4}}{\frac{x^2}{4}+1} dx \\ \frac{1}{4} \int\limits x^{-2} dx-\frac{1}{4} \cdot \frac{1}{4} \int\limits \frac{1}{(\frac{x}{2})^2+1} dx \\ \frac{1}{4 } \int\limits x^{-2} -\frac{1}{16} \int\limits \frac{1}{(\frac{x}{2})^2+1} dx\]
anonymous
  • anonymous
coool. i think is got it from here. i should be able to integrate . i was just confused on the decomposition part. thank you!
freckles
  • freckles
np
anonymous
  • anonymous
i got it from here*

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