Question

how would you separate 4x+16/(x(x-2)^2) into the A, B, and C form? (partial fraction integration)

Answer 1

There's certain rules you have to follow, but for this is
\[\frac{4x+16}{x(x-2)^2 }=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2}\]

Answer 2

when would you have to put Cx+D, instead of just a C over something?

Answer 3

If the denominator contains x power 2, eg.
\[\frac{4x+16}{x(x^2-2) }=\frac{A}{x}+\frac{Bx+C}{x^2-2}\]

Answer 4

If the denominator is quadratic, you use Cx+D.

Answer 5

Not \(x^2-2\), though, that you would factor into \((x+\sqrt2)(x-\sqrt2)\).

Answer 6

Example of a non-reducible quadratic denominator would be \(x^2+1\).

Answer 7

That's as an example, using his question , he'll get the idea

Answer 8

It's a wrong example.

Answer 9

@JSC253 Have you been able to solve for A, B, and C?

Answer 10

for the previous question, yeah. but now im doing the integral of (4x+16)/(x^3-4x^2+4x). i separated it into 4x+16=A(x-2)^2+B(x)(x+2)+(Cx+D)(x). found A and trying to figure out B, C, and D.

Answer 11

could you help me out on this one as well?

Answer 12

Close this question and open a new one :) Also it's nice when you close a question to select a best answer for the question.

Answer 13

Factor the denominator x(x-2)^2
\[\frac{16+4 x}{(-2+x)^2 x}=\frac{a }{x-2}+\frac{b }{(x-2)^2}+\frac{c}{x}\]

Answer 14

Once you've worked out you should get
\[\begin{array}{l} a=-4 \\b=12 \\c=4 \\\end{array}\]