Question

Compute the integral.

Answer 1

You can change the exponents to negatives to move the denominator to the numerator.

Answer 2

partial fractions should work

Answer 3

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

Answer 4

Yeah i got that , having difficulties finding A,B,C, and D

Answer 5

which method are you used to:
heaviside or comparing and solving a system of linear equations?

Answer 6

\[Ax(x^2+4)+B(x^2+4)+(Cx+D)(x^2)=1 \]

Answer 7

both were taught i suppose. i dont mind expanding everything

Answer 8

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
\[(C \cdot (-2i)+D)(4i^2)=1 \\ -2i C+D=\frac{-1}{4} \\ 2iC+D=\frac{-1}{4} \text{ which was given by the other x value we plugged \in } \\ \text{ now add } 2D=\frac{-2}{4}=\frac{-1}{2} \\ \text{ so } D=\frac{-1}{4} \\ \text{ so now } C \\ -2i C-\frac{1}{4}=\frac{-1}{4} \\ -2i C=0 \\ \text{ so we have } C=0\]

Answer 9

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

Answer 10

it's definately not the heaviside lol

Answer 11

I only plugged in -2i and 2i
because it made evaluating x^2+4 simple

Answer 12

since (2i)^2+4 is 0
and (-2i)^2+4 is also 0

Answer 13

ok you expand and I will expand and regroup terms and you can compare my equation to your equation

Answer 14

\[Ax(x^2+4)+B(x^2+4)+(Cx+D)(x^2)=1 \\ A(x^3+4x)+B(x^2+4)+Cx^3+Dx^2 =1 \\ x^3(A+C)+x^2(B+D)+x(4A)+(4B)=1\]