Question

how to solve integral 1/(x^2-x-2)dx ?
wolframalpha solves it weird way. Im looking for way to solve it without making it squares

Answer 1

Use partial fractions

Answer 2

\[\Large \int\limits\limits\limits\frac{1}{x^2-x-2}dx\quad=\quad \int\limits\limits\frac{1}{(x-2)(x+1)}dx\]See how it factors?
Looks like we'll need to do partial fractions.

Answer 3

Remember how to do those? :)

Answer 4

I tried it with A/(x-2) + B/(x+1) but I get A and B as 0. I think I set the equation wrong

Answer 5

\[\Large \frac{1}{(x-2)(x+1)}\quad=\quad \frac{A}{(x-2)}+\frac{B}{(x+1)}\]Multiplying through by the denominator on the left,\[\Large 1\quad=\quad A(x+1)+B(x-2)\]

Answer 6

You can plug in specific x values to help solve for A and B.

If we let x=2, we have the equation:\[\Large 1\quad=\quad A(2+1)+B(2-2)\]

Answer 7

ohh 1/3 forB and -1/3 forA

Answer 8

Mmmm the other way around I think :)

Answer 9

(1/3)ln(x-2)-(1/3)ln(x+1) +C
so (1/3)ln((x-2)/(x+1))+c
great!

Answer 10

yay gj