Can someone explain me the steps of solving integrals using partial fractions?

- anonymous

Can someone explain me the steps of solving integrals using partial fractions?

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- schrodinger

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- anonymous

Should i give you am example?

- anonymous

the integration is usually the easy part. finding the decomposition is a pain often

- anonymous

ya i don't get the steps. my book didnt tell me

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## More answers

- anonymous

|dw:1326817894193:dw|

- anonymous

\[\frac{1}{(x-2)(x-3)}=\frac{a}{x-2}+\frac{b}{x-3}\] and you need a and b. we know that
\[a(x-3)+b(x-2)=1\]
so this is an easy one. since this equality has to be true for all values of x, it must be true if x = 2, so you get
\[a(2-3)+b(2-2)=1\]
\[-a=1\]
\[a=-1\]

- anonymous

lol they got a diff answer is that ok?

- anonymous

How about using heavy side cover up for this one?

- anonymous

repeat the process making x = 3, and you will find
\[b=1\] so we have
\[\frac{1}{(x-2)(x-3)}=\frac{-1}{x-2}+\frac{1}{x-3}\] and now integrate term by term.
now it is not ok

- anonymous

*no

- anonymous

They got 1/3 and -1/3

- anonymous

then maybe i made a mistake, let me write it out

- TuringTest

is that a 5 on the second set of parentheses?

- TuringTest

sat used a 3

- anonymous

doh, i used 3!!

- anonymous

LOL ya sorry abt that

- anonymous

ohhh didnt even notce that sat

- TuringTest

sat has done it correctly for sure, just with the wrong number

- anonymous

no that is my fault, but idea is clear right?

- anonymous

\[a(x-5)+b(x-2)=1\] let
\[x=2\] get
\[-3a=1,x=-\frac{1}{3}\]

- anonymous

ummmm so y did u chose the value 2?

- anonymous

i am going to wager you can figure out why i picked 2

- anonymous

and what i will pick next also

- anonymous

need a hint?

- anonymous

well cuz it is in the denomanator? lol

- anonymous

i have this equality
\[a(x-5)+b(x-2)=1\] and i am looking for a and b. so how can i find a easily?

- anonymous

by eliminating one variable by making one part =0

- anonymous

K i got that LOL

- anonymous

zactly.
so first we let x =2, and then we let x = 5

- anonymous

oh ok got that :D Thanks :D

- anonymous

yw

- anonymous

I thought u guys started teaching today

- anonymous

This one a is a simple one if I have a more difficult one I will be back :D

- anonymous

with some practice you can do this: write
\[\frac{1}{(x-2)(x-5)}=\frac{a}{x-2}+\frac{b}{x-5}\] now to find a, put your finger over the x - 2 in the first expression, put x = 2 and get
\[a=\frac{1}{2-3}=-\frac{1}{3}\]

- myininaya

its called office hours

- anonymous

Thanks sat :D

- anonymous

i really need to get back to work, but i have 30 emails to send out and got bored bored bored.
later

- anonymous

LOL bye

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