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?

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.