Question

Another Partial Functions Problem

Answer 1

please give me second to upload the problem and my work

Answer 2

\[\int\limits_{}^{}\frac{ 4 }{ x^2 + 5x - 14 }dx\]

Answer 3

\[\frac{A}{x+7}+\frac{B}{x-2}\] is a start

Answer 4

\[\int\limits_{}^{}\frac{ 4 }{ x^2 + 5x - 14 }dx = \int\limits_{}^{}\frac{ 4 }{ (x-1)(x+1) }dx = \frac{ A }{ x+7 } + \frac{ B }{ x-2 }\]

Answer 5

you want to find A and B the snappy way?

Answer 6

or the slower way (same thing really)

Answer 7

I was told to find A and B by doing system of equations. I do know that you can also find A and B quickly by letting x = -7 and x = 2, then solving for A and B. But let's do it the first way because that's where I messed up

Answer 8

lets do it the real snappy way first

Answer 9

Okay

Answer 10

The way I did it:
4 = A(x-2) + B(x+7)
4 = Ax - 2A + Bx + 7B
4 = (A+B)x - 2A + 7B

4 = -2A - 7B A + B = 0
4 = -2(-B) - 7B A = -B
4 = 2B - 7B A = -(-4/5)
4 = -5B A = 4/5
-4/5 = B

Answer 11

\[\frac{4}{(x+7)(x-2)}=\frac{A}{x+7}+\frac{B}{x-2}\] to find \(A\) put your finger over the factor of \(x+7\) in \[\frac{4}{(x+7)(x-2)}\]and replace \(x\) by \(-7\) \[\frac{4}{\cancel{(x+7)}(-7-2)}\]

Answer 12

you get \[A=-\frac{4}{9}\] instantly

Answer 13

you want to do it your "system of equations" long boring way and see if we can find the mistake?

Answer 14

Then,
\[\int\limits_{}^{}\frac{ 4 }{ x^2 + 5x - 14 }dx = \int\limits_{}^{}\frac{ 4 }{ 5(x+7) }dx - \int\limits_{}^{}\frac{ 4 }{5(x-2) }dx\]

So the answer I got is \[= \frac{ 4 }{ 5 }\ln |x+7| - \frac{ 4 }{ 5 }\ln |x-2| + C\]

Answer 15

yeah the 5 is wrong, should be 9

Answer 16

I understand the method you used! Yes, can we try the boring way? Lol. I just want to make sure I can do both methods

Answer 17

Yeah, I don't know where I went wrong. Somehow I ended up with 5 instead of 9

Answer 18

ok i like the "put your finger over the factor" method a lot more

Answer 19

Haha, me too! But I need to be familiar with both methods for my exam

Answer 20

ok sure lets see the mistake
\[A(x-2)+B(x+7)=4\]is a start
then \\[Ax-2A+Bx+7B=4\]so \[A+B=0\\-2A+7B=4\]

Answer 21

Oh wait lol i see it

Answer 22

oh i see it right away
since \(A=-B\) you have \[2B+7B=4\\
B=\frac{4}{9}\]

Answer 23

I wrote 4 = -2A - 7B instead of 4 = -2A + 7B.

Answer 24

ooh ok well glad we cleared that up!

Answer 25

then A = -4/9 and B = 4/9 and then we get the right answer haha. Thank you!

Answer 26

yw

Answer 27

there was a fellow here a while ago "eliasaab" i think, who had an amazingly snappy way to do this, even if you had annoying repeated roots
i forget what it was though, had to do with derivatives

Answer 28

nice, maybe i'll come across the method again