Question

\[\int_{3}^{4} \frac{x^3 -2x^2 -4}{x^3 -2x^2}dx\]

Answer 1

\[\int_{3}^{4} \frac{x^3 -2x^2 -4}{x^3 -2x^2}dx\]
\[=\int_{3}^{4} 1-\frac{4}{x^3 -2x^2}dx\]

Answer 2

this is how far I've gotten, not sure if it's right
\[\int_{3}^{4} 1+\frac{1}{x}-\frac{3}{x-2}dx\]

Answer 3

\[\frac{-4}{x^3-2x^2}=\frac{-4}{x^2(x-2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-2}\]

Answer 4

By the way you can always check yourself by combining you fractions to see if you ended up with what you started out with
But anyways I think you are missing x^2 on bottom somewhere

Answer 5

Which means you broke it down wrong

Answer 6

yep that's what I did...no actually
\[1+\frac{Ax}{x^2}+\frac{B}{x-2}\]

Answer 7

Yeah you are missing the over x^2 part

Answer 8

where?

Answer 9

Look what I wrote for you above
Find constants A,B,and C such that above is true

Answer 10

I had to do long division first because the numerator was equal to the denominator and I got \[1+\frac{4}{x^3 -2x^2}\]

Answer 11

i mean the powers

Answer 12

x^3

Answer 13

Right

Answer 14

Now you need to find
And I didn't do the long division part so I don't know who is right about you or Unkle
But anyways the bottom factors like this x^2(x-2)
So that means you must do
\[\frac{4}{x^2(x-2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-2}\]

Answer 15

Actually Unkle is right
You don't even need long division for this

Answer 16

so it wouldn't be \[\frac{Ax}{x^2}\]
but instead
\[\frac{A}{x}+\frac{B}{x^2}\]
?

Answer 17

\[\frac{x^3-2x^2-4}{x^3-2x^2}=\frac{x^3-2x^2}{x^3-2x^2}+\frac{-4}{x^3-2x^2}\]
\[1+\frac{-4}{x^3-2x^2}\]

Answer 18

ok

Answer 19

Yes it is what I said I'm 100 percent sure

Answer 20

Try combining what you did, do you get what you started with
The answer is no

Answer 21

all right

Answer 22

thanks @myininaya !

Answer 23

by combining you mean...?

Answer 24

just adding the terms and seeing if I get what I started with?

Answer 25

Oh come on you know how to combine fractions :p
For example to combine the following fractions (to perform the addition we need to find a common denominator)
\[\frac{5}{x-2}+\frac{4}{x}\]
\[\frac{5x}{x(x-2)}+\frac{4(x-2)}{x(x-2)}=\frac{5x+4(x-2)}{x(x-2)}\]
Example if we wanted to do partial fractions on: Just like if we had
\[\frac{4}{x(x-2)^2}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2}\]