Question

evaluate the definite integral: F 2at the top and 1at the bottom (3/x^2 - 1) dx

Answer 1

Hi,
can you do partial fractions ? or know what that is ?

Answer 2

what is partial fractions?

Answer 3

It is a way of "breaking apart" fractions with polynomials in them.

Answer 4

split the denom as (x-1)(x+1) the the numer as 3[(x+1) - (x-1)]/2 then separate out the num... and no partial fractions will be needed

Answer 5

yea we can use partial fractions

Answer 6

something like this :
\(\Large \dfrac{3}{x^2-1}=\dfrac{3}{(x+1)(x-1)}=\dfrac{A}{x-1}+\dfrac{B}{x+1}\)

Answer 7

yea we can use that

Answer 8

do you understand that ? and know how to find A,B?

Answer 9

will they both be 3?

Answer 10

not actually, you can take the LCM right ?

Answer 11

or cross - multiply ?
do it for right side

Answer 12

to have only one common denominator of (x-1)(x+1)

Answer 13

so if you cross multiply what will you get?

Answer 14

can you try once ?
show me what u get and there is error i'll correct it

Answer 15

like a/b +c/d = (ad+bc)/bd

Answer 16

then similarly A/(x-1)+B/(x+1) = ... ?

Answer 17

i am not sure what you mean cause i have never learnt it this way so i am confused but ill send a picture what the question you asked me to do, im not sure if it is correct

Answer 18

how did the x+1 go in the numerator ? thats incorrect...

Answer 19

i'm not sure what to do im really confused right now

Answer 20

you got this right ?
a/b +c/d = (ad+bc)/bd

Answer 21

yea

Answer 22

so, using that
A/(x-1)+B/(x+1) =... ?

Answer 23

(a/(x+1) + (x-1)/B) / (x-1) (x+1) ?

Answer 24

good!

Answer 25

so, you see now that denominators are common for both sides!
so we can equate the numerators
we get
3 = A(x+1) +B(x-1)
clear till here ?

Answer 26

ohh yeaa i get that now

Answer 27

good, so that equation is true for every value of x
put x= 1 in that equation, what do u get ?

Answer 28

3=A

Answer 29

are you sure ?

Answer 30

3 = A (1+1)
A = ...?

Answer 31

6

Answer 32

noo 3/2

Answer 33

yes

Answer 34

now to find B, you put x =-1 in that equation, what do u get ?

Answer 35

-3/2

Answer 36

yes!
so did you get how we found A and B ?
and
\(\Large \dfrac{3}{(x+1)(x-1)}=\dfrac{A}{x-1}+\dfrac{B}{x+1}=\dfrac{3}{2}[\dfrac{1}{x-1}-\dfrac{1}{x+1}]\)
?

Answer 37

yes i get that now

Answer 38

and can you now integrate those 2 terms separately ?

Answer 39

so do i now replace the x with 2 first and then take away what i get when i replace x with 1?

Answer 40

but you need to integrate first right ?
then you'll do that

Answer 41

remember we still haven't integrated,
you know the integration for 1/(x-a) general formula ?

Answer 42

lnx-a right?

Answer 43

yes!
so integral of 1/(x-1) = ... ?
so integral of 1/(x+1) = ... ?

Answer 44

lnx-1 - lnx+1

Answer 45

so the whole thing will be 3/2 x (lnx-1 - lnx+1)

Answer 46

x not times

Answer 47

yes,
3/2 (ln |x-1| - ln |x+1|)
now you can do what you asked,
"so do i now replace the x with 2 first and then take away what i get when i replace x with 1?
"

Answer 48

ohh ok i get it now thanks alot!

Answer 49

so whats your final answer ?

Answer 50

-0.61 ?

Answer 51

no...when you put x= 1
you get ln |0| right ?

Answer 52

yea so is it equal to zero?

Answer 53

nopes, actually ln 0 = -infinity
so your final answer will be +infinity
got this ?

Answer 54

ohh okk i see thankss again

Answer 55

oh and since you are new here,
\(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)
if you have any doubts browsing this site, you can ask me.
Have fun learning with Open Study! :)

Answer 56

ohh okk thank you!! :)

Answer 57

welcome ^_^