Question

Calculus 2 - partial fractions

Answer 1

\[\frac{32}{y(y-4)(y+4)}=\frac{A}{y}+\frac{B}{y-4}+\frac{C}{y+4}\]

Answer 2

put \(y=0\) get
\[A=\frac{32}{-4\times 4}=-4\]

Answer 3

put \(y=4\) get
\[B=\frac{32}{4\times 8}=1\]

Answer 4

put \(y=-4\) get \[C=\frac{32}{-4\times -8}=1\]

Answer 5

\[\frac{32}{y(y-4)(y+4)}=\frac{2}{y}+\frac{1}{y-4}+\frac{1}{y+4}\] lets check it

Answer 6

god i am an idiot tonight
put \(y=0\) get \(A=\frac{32}{-4\times -4}=-2\)

Answer 7

\[\frac{32}{y(y-4)(y+4)}=-\frac{2}{y}+\frac{1}{y-4}+\frac{1}{y+4}\]

Answer 8

Do you type the answer in that form?

Answer 9

that is the partial fraction decomposition
now you can integrate pretty much in your head
\[-2\ln(y)+\ln(y-4)+\ln(y+4)\]

Answer 10

ohh ok, thank you