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anonymous
 one year ago
Compute the integral.
anonymous
 one year ago
Compute the integral.

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tacotime
 one year ago
Best ResponseYou've already chosen the best response.0You can change the exponents to negatives to move the denominator to the numerator.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1partial fractions should work

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{1}{x^2(x^2+4)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+4}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah i got that , having difficulties finding A,B,C, and D

freckles
 one year ago
Best ResponseYou've already chosen the best response.1which method are you used to: heaviside or comparing and solving a system of linear equations?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[Ax(x^2+4)+B(x^2+4)+(Cx+D)(x^2)=1 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0both were taught i suppose. i dont mind expanding everything

freckles
 one year ago
Best ResponseYou've already chosen the best response.1for 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
 one year ago
Best ResponseYou've already chosen the best response.1but 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
 one year ago
Best ResponseYou've already chosen the best response.0it's definately not the heaviside lol

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I only plugged in 2i and 2i because it made evaluating x^2+4 simple

freckles
 one year ago
Best ResponseYou've already chosen the best response.1since (2i)^2+4 is 0 and (2i)^2+4 is also 0

freckles
 one year ago
Best ResponseYou've already chosen the best response.1ok you expand and I will expand and regroup terms and you can compare my equation to your equation

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[Ax(x^2+4)+B(x^2+4)+(Cx+D)(x^2)=1 \\ A(x^3+4x)+B(x^2+4)+Cx^3+Dx^2 =1 \\ x^3(A+C)+x^2(B+D)+x(4A)+(4B)=1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444155517948:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.1great 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
 one year ago
Best ResponseYou've already chosen the best response.1I would solve the last two equations first because they only involve one constant variable each

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[A=0 , B=\frac{1}{4}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1and then go to the top two equations and used these values to solve for C and D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hold 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
 one year ago
Best ResponseYou've already chosen the best response.1\[Ax(x^2+4)+B(x^2+4)+(Cx+D)(x^2)=1 \\ A(x^3+4x)+B(x^2+4)+Cx^3+Dx^2 =1 \\ x^3(A+C)+x^2(B+D)+x(4A)+(4B)=1 \text{ same as } \\ x^3(A+C)+x^2(B+D)+x(4A)+(4B)=x^3(0)+x^2(0)+x(0)+1\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1we 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
 one year ago
Best ResponseYou've already chosen the best response.0A=0 B=1/4 C=0 D=1/4

freckles
 one year ago
Best ResponseYou've already chosen the best response.1sounds 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{1}{4} \int\limits x^{2} dx\frac{1}{4} \int\limits \frac{1}{x^2+4} dx \\ \frac{1}{4} \int\limits x^{2} dx  \frac{1}{4 } \int\limits \frac{\frac{1}{4}}{\frac{x^2}{4}+1} dx \\ \frac{1}{4} \int\limits x^{2} dx\frac{1}{4} \cdot \frac{1}{4} \int\limits \frac{1}{(\frac{x}{2})^2+1} dx \\ \frac{1}{4 } \int\limits x^{2} \frac{1}{16} \int\limits \frac{1}{(\frac{x}{2})^2+1} dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0coool. i think is got it from here. i should be able to integrate . i was just confused on the decomposition part. thank you!
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