Question

Calculus 2,
http://screencast.com/t/pECz6sym
O_O :)

I hope you understand my question.

Answer 1

what's the gcd of that expression? that's why you have to take out the x^2 term

Answer 2

I don't know whats gcd

Answer 3

greatest common divisor

Answer 4

hm

Answer 5

the greatest common divisor , isn't that x^2 - 4 ?

Answer 6

oh wait

Answer 7

you have x^3 - x^2 the gcd is x^2

Answer 8

I don't quite get it to be honest

Answer 9

because it has an even power ? or that's irrelevant ?

Answer 10

this is a partial fraction decomposition, correct?

Answer 11

yes

Answer 12

the whole idea here is to rewrite the fraction as a sum (or difference) of fractions with denominators whose terms are linear or irreducible quadratic functions.
then you can integrate term by term.

anytime yu have a repeated factor, like x^2, this reduces to \[\frac{ A }{ x }+\frac{ B }{ x^2 }=\frac{ Ax+B }{ x^2 }\]

if you don't account for what's possible, you'll not get the correct answer

Answer 13

if the denominator is an irreducible quadratc, then you must have a linear term in the numerator...
\[\frac{ Ax+B }{ x^2 +1 }\]

Answer 14

ok sorry for asking many questions but can we define linear ? I don't even understand it in my mother language

Answer 15

and irreducible quadratic

Answer 16

linear equation... like y = mx+b

irreducible => the equation has complex roots

Answer 17

how do I know if an equation has complex roots ? By complex roots you meant complex numbers as a result ?

y = mx+b so linear is an equation of a line ?

Answer 18

if
\[ax^2+bx+c\]is your quadratic, then if
\[b^2-4ac<0\]
then the quadratic is irreducible and has complex roots

Answer 19

mx+b, or similarly, Ax + B is linear in the variable x because x has a power of 1 in the expression.

Answer 20

So I don't have a linear term in my numerator ? Why ?

Answer 21

let's look at your example a little closer...

\[\frac{ 2x-3 }{ x^3 - x^2 }=\frac{ A }{ x }+\frac{ B }{ x^2 }+\frac{ C }{ x-1 }=\frac{ Ax(x-1) }{ x\cdot x(x-1) }+\frac{ B(x-1) }{ x^2(x-1) }+\frac{ Cx^2 }{ x^2(x-1) }\]

The denominator had to be broken up in that way to account for all possibilities. Otherwise, you'd end up with an incorrect solution or an insolvable set of equations.

in the last expression, everything is to be multiplied out\[=\frac{ Ax(x-1) }{ x\cdot x(x-1) }+\frac{ B(x-1) }{ x^2(x-1) }+\frac{ Cx^2 }{ x^2(x-1) }=\frac{ Ax^2-Ax }{ x^3-x^2 }+\frac{ Bx-B }{ x^3-x^2 }+\frac{ Cx^2 }{ x^3-x^2 }\]
\[=\frac{(A+C)x^2+(B-A)x+B) }{ x^3-x^2 }\]

since the original numerator was 2x-3, this implies that A+C = 0 since ther is no x^2 term in the origianl numerator.

also, B-A = 2 since the x term has a coefficient of 2.

likewise, B = -3 since the it is the only constant term and the constant term in the original numerator was -3.

From that we can deduce that A = -5 and C = 5, since -3-A = 2 and since A+C = 0.

thus we get the decomposition:

\[=\frac{-5 }{x }+\frac{ -3 }{ x^2 }+\frac{ 5 }{ x-1 }=\frac{ 5 }{ x-1 }-\frac{5 }{x }-\frac{ 3 }{ x^2 }\]

Answer 22

@Christos please have a look