Question

Help Partial Fractions
\[\int\limits_{}^{}\frac{ dx }{ (x+1)(x+2) }\]

Answer 1

Any confusion on that setup or multiplication?

Answer 2

no but before we move on I actually meant to put x-1

Answer 3

just to clear things up

Answer 4

\[\Large \color{teal}{\frac{1}{(x-1)(x+2)}\quad=\quad \frac{A}{x-1}+\frac{B}{x+2}}\]Oh ok :) I'm gonna erase the other comment then so we don't get confused.\[\Large \color{teal}{1\quad=\quad A(x+2)+B(x-1)}\]

Answer 5

ok

Answer 6

To solve for the unknown constants A and B, we have a couple of options:
~We can multiply out the brackets and equate like terms ( which ends up being really messy in a lot of cases, not this one though ).
~Or we can plug in specific x values that will cancel out one of the unknowns, allowing us to solve for the other.

Answer 7

For example: if we plug in \(\Large x=1\),
what does that give us here?\[\Large \color{teal}{1\quad=\quad A(x+2)+B(x-1)}\]

Answer 8

a=1/3

Answer 9

It cancelled out the B, allowing us to solve for A.
Cool!

So what value x-value would allow us to cancel the A, and solve for B? :X

Answer 10

-2

Answer 11

sounds good c:

Answer 12

b=-1/3

Answer 13

Ok good, so here is what we've determined.\[\Large \frac{A}{x-1}+\frac{B}{x+2}\quad=\quad \frac{(1/3)}{x-1}+\frac{(-1/3)}{x+2}\]It's always ugly writing fractions on fractions -_- lol

Answer 14

ok

Answer 15

\[\large \int\limits\frac{1}{(x-1)(x+2)}dx\quad=\quad \int\limits \frac{A}{x-1}+\frac{B}{x+2}\;dx\]
\[\large =\quad \int\limits \frac{(1/3)}{x-1}-\frac{(1/3)}{x+2}\;dx\]

Answer 16

This will be easier to read if we factor a 1/3 out of each term and pull it outside of the integral.\[\Large =\quad \frac{1}{3}\int\limits \frac{1}{x-1}-\frac{1}{x+2}\;dx\]

Answer 17

Understand how to solve it from here? :o

Answer 18

so \[(\frac{ 1 }{ 3 })\ln \frac{ x+2 }{ x-1 }\]

Answer 19

+C

Answer 20

Hmm I think your log stuff is upside-down, no?

Answer 21

\[\Large \log(a)-\log(b)\quad=\quad \log\left(\frac{a}{b}\right)\]

Answer 22

oh x-2 is negative my fault not x-1

Answer 23

alright thanks i think I can handle this

Answer 24

yay team \c:/