Question

Partial Fractions\[\int\limits_{}^{}\frac{ dx }{ x^2(x^2-16) }\]

Answer 1

so far i have
A/(x)+B/(x^2)+C/(x+4)+D/(x-4)

Answer 2

Yep that's right

Answer 3

then i multiplied and simplified basically there's alot more to right out but i cant seem to solve for A or B

Answer 4

If you have C and D you can sub in two arbitrary x values that aren't -4 or 4 and you can build a system of equations

Answer 5

but wouldn't that leave you with two things to solve for or am i misunderstanding you

Answer 6

\[\frac{1}{x^2(x^2-16)}=\frac{Ax(x+4)(x-4)+B(x+4)(x-4)+Cx^2(x+4)+Dx^2(x-4)}{x^2(x^2-16)}\]
\[1=x^3(A+C+D)+x^2(B+4C-4D)+x(-16A)+(-16B)\]
A and B should be easiest to solve for unless I messed up somewhere

Answer 7

sub x = 4, you find C. sub x = -4, you find D. Next you can sub x = 1 and x = 2, that gives you two equations with two unknowns you can solve.

Answer 8

So you are suppose to have
\[1=x^3(0)+x^2(0)+x(0)+1\] (so we can have 1=1)
and you have
\[1=x^3(A+C+D)+x^2(B+4C-4D)+x(-16A)+(-16B)\]
You have 4 equations to solve.
A+C+D=0
B+4C-4D=0
-16A=0
-16B=1
These easiest two equations to solve are the last two since there is only one unknown in each.

Answer 9

ok ty