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

- megannicole51

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

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

- anonymous

is this partial fractions?

- anonymous

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

- anonymous

that is what you are looking for, right?

Looking for something else?

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

## More answers

- anonymous

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
\[\frac{4}{4-5}=-4\]

- megannicole51

yes it is....sorry my laptop died

- anonymous

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

- anonymous

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

- megannicole51

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

- anonymous

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

- anonymous

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

- anonymous

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

- megannicole51

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

- anonymous

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

- megannicole51

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

- anonymous

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

- anonymous

@satellite73 you show off...

- anonymous

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

- megannicole51

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:)

- anonymous

that is not correct though

- anonymous

like i said... you get 2 equations in 2 unknowns...
A + B = 1
-4A-5B=0
solve the system

- anonymous

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

- megannicole51

why is A+B=1?

- anonymous

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

- anonymous

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

- anonymous

i know

- anonymous

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

- anonymous

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

- megannicole51

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

- anonymous

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

- megannicole51

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

- anonymous

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

- anonymous

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

- anonymous

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

- megannicole51

oooooh that makes sense

- megannicole51

okay let me see if i can finish it by myself

- anonymous

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

- megannicole51

let me do it real fast then okay!

- anonymous

k

- anonymous

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

- megannicole51

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

- megannicole51

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

- anonymous

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

- megannicole51

oh okay

- anonymous

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

- megannicole51

gotcha!

- megannicole51

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

- anonymous

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

- megannicole51

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

- megannicole51

im doing substitution

- anonymous

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

- anonymous

clear or no?

- megannicole51

yeah it is hold on

- megannicole51

im having pencil issues lol

- anonymous

k

- megannicole51

A=5
B=-4

- anonymous

whew!!

- megannicole51

yay!

- anonymous

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

- megannicole51

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

- anonymous

then lets to it the snappy way because this is ridiculous

- megannicole51

hahah okay show me!

- anonymous

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

- megannicole51

yes

- megannicole51

no

- megannicole51

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

- anonymous

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

- megannicole51

lol yeah bt keep goin!

- anonymous

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

- megannicole51

:) good

- anonymous

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

- megannicole51

yup

- anonymous

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

- anonymous

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

- anonymous

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

- megannicole51

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

- megannicole51

dang! i like it:)

- megannicole51

thank you for showing me!

- anonymous

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

- megannicole51

i love math lol

- anonymous

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

- megannicole51

just find the zeros?

- anonymous

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

- anonymous

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 \(5\)

- anonymous

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

- anonymous

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

- megannicole51

lol agreed but thts wat i love about math

- anonymous

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

- megannicole51

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

- anonymous

yw

Looking for something else?

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