Question

evaluate the integral (dy/(4-y^2))

Answer 1

\[\frac{1}{4-y^2} = \frac{1}{(2-y)(2+y)}\]
now break into partial fractions

Answer 2

i was kinda partial to it too

Answer 3

\[\frac{1}{4-y^2} = \frac{A}{2-y}+\frac{B}{2+y}\]\[1=A(2+y)+B(2-y)\]\[1=2A+Ay+2B-By\]\[2A+2B=1\]\[A-B=0\]\[A=B\]\[4A=1\]\[A=B=\frac{1}{4}\]
\[\int\limits{\frac{dy}{4-y^2}}=\int\limits{\frac{dy}{4(2-y)}}+\int\limits{\frac{dy}{4(2+y)}}\]

Answer 4

can u integrate now?

Answer 5

well couldnt we just use substitution

Answer 6

lol .... and sub into what? y^2 doesnt derive into a constant

Answer 7

no i mean using u substitution ?

Answer 8

partial fractions my favorite! you know how to do it the french way?

Answer 9

\[\frac{1}{4} \int\limits{\frac{dy}{2-y}}+\frac{1}{4}\int\limits{\frac{dy}{2+y}} = \frac{1}{4} (\ln(2-y)+\ln(2+y))\]

Answer 10

you want
\[\frac{1}{(2-x)(2+x)}=\frac{A}{2-x}+\frac{B}{2+x}\] so find A think " x cannot be 2} to take
\[\frac{1}{(2-x)(2+x)}\] get rid of the factor x-2 and put
\[\frac{1}{2+x}\] replace x by 2 and get
\[A=\frac{1}{4}\]

Answer 11

you can do it in your head by putting your finger over the factor you don't want and substituting in the number. french guy showed me that easy trick

Answer 12

i like it :D