Question

evaluate ∫1/(x-2)(x^2+1) using partial fractions

Answer 1

i have it set up like this...

A/(x-1) +Bx/(x^2+1) +C/(x^2+1)

Answer 2

\[\frac{A}{x-1}+\frac{Bx+C}{x^2+1}\] is easier

Answer 3

our professor told us to break it up further...

Answer 4

don't listen
you can break it apart later if you need to

Answer 5

alright then lets do it your way lol

Answer 6

wouldnt it be (Bx+C)(x-2) ?

Answer 7

ok i misread the whole thing
it is
\[\int \frac{dx}{(x-2)(x^2+1)}\] right?

Answer 8

correct

Answer 9

ok then
\[A(x^2+1)+(Bx+C)(x-2)=1\] that's better

Answer 10

yeah

Answer 11

now put \(x=2\)

Answer 12

so A would = 5?

Answer 13

you get \(5A=1\) making \(A=\frac{1}{5}\)

Answer 14

To check the final result
https://www.wolframalpha.com/input/?i=partial+fraction+1%2F%28%28x-2%29%28x^2%2B1%29%29

Answer 15

you get \(B=-\frac{1}{5}\) right away since you have \(Ax^2+Bx^2=0\) and you know \(A=\frac{1}{5}\)
let me know if that is obvious or not

Answer 16

kinda

Answer 17

mentally multiply this out in your head, don't really multiply it
\[A(x^2+1)+(Bx+C)(x-2)=1\]
where will you get an \(x^2\) term?
one will be from the first term \(Ax^2\) and the other will be the \(Bx^2\) from the second term
that is the only place you will get an \(x^2\)
so on the left you have \(Ax^2+Bx^2\) and on the right you have no \(x^2\) meaning
\[A+B=0\] and since \(A=\frac{1}{5}\) then \(B=-\frac{1}{5}\)

Answer 18

of course you are free to multiply all that much out, combine like terms, and then equate like coefficients
and then solve an annoying system of equations
i would look for shortcuts

Answer 19

similarly, where is the constant going to come from?
the \(A\times 1\) in the first term, and the \(-2C\) in the second term
so you know \(A-2C=1\) and again since \(A=\frac{1}{5}\) you have \(C=-\frac{2}{5}\)

Answer 20

ohhh. okay gotcha. then you sub those values back into the broken up integral and then solve right?

Answer 21

yes,
check the wolf answer for the nice way to write it
i would actually write
\[\frac{1}{5}\int \frac{dx}{x-5}-\frac{1}{5}\int\frac{x+2}{x^2+1}dx\]

Answer 22

first is a log, second is a simple u-sub

Answer 23

okay thanks

Answer 24

yw