Question

solve the integral ((x)/(x^2-9x+20))dx

I solved it all the way to the bottom and now i am at x(A+B)+A(-4)+B(-5)=x

I was told that A+B can only equal 0 when the number after the equal sign (in this case "x" cannot be a variable). So i tried solving it out anyways and got A=x and B=-x. Then I plugged it into the original ((A)/(x-5))+((B)/(x-4)) and it doesn't work.

@pgpilot326

Answer 1

is this partial fractions?

Answer 2

\[x^2-9x+20=(x-4)(x-5)\]
\[\frac{x}{(x-4)(x-5)}=\frac{A}{x-4}+\frac{b}{x-5}\] there is a real quick snappy way to solve

Answer 3

that is what you are looking for, right?

Answer 4

A and B are numbers, not variables.
one way to find A is this
take \[\frac{x}{(x-4)(x-5)}\] put your hand over the part of the denominator that is the denominator for A \[\frac{x}{\cancel{(x-4)}(x-5)}\] replace \(x\) by \(4\) and get
\[\frac{4}{4-5}=-4\]

Answer 5

yes it is....sorry my laptop died

Answer 6

ok so now we have
\[\frac{-4}{x-4}\] for the A part

Answer 7

\[ \frac{x}{x^{2}-9x+20}=\frac{A}{x-5}+\frac{B}{x-4} \Rightarrow A(x-4)+B(x-5)=x\]

Answer 8

yup ive got that....then i have x(A+B)+A(-4)+B(-5)=x

Answer 9

for B repeat and get
\[\frac{x}{(x-4)\cancel{(x-5)}}\]

Answer 10

you get 2 equations in 2 unknowns... this is a longer way as it should yield the same thing @satellite73 was saying.

Answer 11

put in
\(x=5\) to get
\[\frac{5}{5-4}=5\]

Answer 12

@satellite73 i dont understand why you are crossing them out of the denominator

Answer 13

ok lets go back to the beginning
i was just showing off a snappy trick

Answer 14

so @pgpilot326 what do i do after..x(A+B)+A(-4)+B(-5)=x

Answer 15

\[A(x-5)+B(x-4)=x\] replace \(x\) by \(5\)

Answer 16

@satellite73 you show off...

Answer 17

you get
\[B(5-4)=5\] sp
\[B=5\]

Answer 18

i dont need to go to the beginning i just need to know what to do after x(A+B)+A(-4)+B(-5)=x please:)

Answer 19

that is not correct though

Answer 20

like i said... you get 2 equations in 2 unknowns...

A + B = 1
-4A-5B=0

solve the system

Answer 21

\[\frac{x}{x^{2}-9x+20}=\frac{A}{x-5}+\frac{B}{x-4}\] so
\[A(x-4)+B(x-5)=x\] for the numerator
now this has to be true for all \(x\)
in particular it must be true of \(x=5\)

Answer 22

why is A+B=1?

Answer 23

Ax+Bx=x so this is the same as A+B=1

Answer 24

i will shut up except to say this is a long and hard way to do it

Answer 25

i know

Answer 26

would you like to do it the easy way? without solving a system of equations?

Answer 27

the main thing is that @megannicole51 understand and can do it herself... it doesn't matter to me

Answer 28

i just need to know how to do it after x(A+B)+A(-4)+B(-5)=x.....ive been doing them like this all day so i dont want to get confused

Answer 29

ok then as @pgpilot326 said, this means \(A+B=1\) since \((A+B)x=1x\) and also that \(-4A-5B=0\) since there is no constant on the right hand side

Answer 30

so if there is no constant A+B will always equal 1 by default?

Answer 31

x terms with x terms and constant terms with constant terms...

Answer 32

no it is because \(x=(A+B)x\)

Answer 33

if you had for example \(3x\) on the right, then \(A+B=3\)

Answer 34

oooooh that makes sense

Answer 35

okay let me see if i can finish it by myself

Answer 36

ok, then if you like, i can show you how to do it in your head without writing anything down

Answer 37

let me do it real fast then okay!

Answer 38

k

Answer 39

you're in good hands @megannicole51 , I'll leave you be

Answer 40

thank you for your help ill message you if i need help! :D

Answer 41

so why does -4A-5B=0?

Answer 42

because A and B are numbers
so \(-4
A-5B\) is a number (constant)
but there is no constant in the numerator, only \(x\)

Answer 43

oh okay

Answer 44

if the numerator was \(x+7\) then \(-4A-5B=7\)

Answer 45

gotcha!

Answer 46

okay so in order to solve for A and B i just do simple algebra and substitute correct?

Answer 47

you solve the system however you like
elimination or in this case substitution which will probably be easier

Answer 48

so i have -4A+5(4A/5)=0....am i on the right track?

Answer 49

im doing substitution

Answer 50

no i don't think so
\[A+B=1\iff B=1-A\] substitute in to
\[-4A-5B=0\] and solve for \(A\)

Answer 51

clear or no?

Answer 52

yeah it is hold on

Answer 53

im having pencil issues lol

Answer 54

k

Answer 55

A=5
B=-4

Answer 56

whew!!

Answer 57

yay!

Answer 58

you see how long that took?
all that work?
multiply out, solve a system yechh

Answer 59

well i have to do it for tests and quizzes and homework :/

Answer 60

then lets to it the snappy way because this is ridiculous

Answer 61

hahah okay show me!

Answer 62

do you have this
\[\frac{x}{x^{2}-9x+20}=\frac{A}{x-4}+\frac{B}{x-5}\] somewhere on your paper?

Answer 63

yes

Answer 64

no

Answer 65

my 4 and 5 are switched but it doesnt really matter does it?

Answer 66

oh ok is it switched? just want to make sure so we don't screw up

Answer 67

lol yeah bt keep goin!

Answer 68

\[\frac{x}{x^{2}-9x+20}=\frac{A}{x-5}+\frac{B}{x-4}\]

Answer 69

:) good

Answer 70

so also on your paper you have
\[A(x-4)+B(x-5)=x\] right?

Answer 71

yup

Answer 72

ok now here is the deal:
\[A(x-4)+B(x-5)=x\] must be true NO MATTER WHAT \(x\) IS !

Answer 73

in particalular it must be true if \(x=5\) and if \(x=5\) you get
\[A(5-4)+B(5-5)=5\] i.e.
\[A=5\]

Answer 74

similarly it must be true if \(x=4\) which, skipping the stupid step of writing \(4-4\) we go right two
\[B(4-5)=4\] or \(-B=4\) and so \(B=-4\) done finished no system etc

Answer 75

yeah thats how you do it when there isnt a coefficient after the equal sign

Answer 76

dang! i like it:)

Answer 77

thank you for showing me!

Answer 78

and there is an even easier way, which amounts to the same thing

Answer 79

i love math lol

Answer 80

\[\frac{x}{(x-5)(x-4)}=\frac{A}{x-5}+\frac{B}{x-4}\]

Answer 81

just find the zeros?

Answer 82

find \(A\)
the denominator of \(A\) is \(x-5\) so replace \(x\) by \(5\)
\[\frac{x}{x-4}\]

Answer 83

that is what i meant by this
\[\frac{x}{\cancel{(x-5)}(x-4)}\] put your hand over the denominator of \(A\) and then replace \(x\) by \(5\)

Answer 84

you get \(A=5\) in your head

Answer 85

much much easier than multiplying out, collecting terms, equating coefficients, solving linear systems, it makes me exhausted just to think about it

Answer 86

lol agreed but thts wat i love about math

Answer 87

glad you love it too
btw the method i just showed obviously doesn't work all the time for example if you are going to end up with something like
\[\frac{2x+1}{x^2+1}+\frac{x-2}{x^2+7}\] however you should always check to see if you can plug in anything for \(x\) to drop out all terms but one, making the solution to one of the constants simple

Answer 88

oh okay! good to know! thank you for your help!!!

Answer 89

yw