Question

Use partial fraction decomposition to find the antiderivative for , i'll post it in a second

Answer 1

\[\frac{ 5x^2+6x+2 }{ (x+2)(x^2+2x+5)}\]

Answer 2

And while we're at it , when I try to find the zeros for x^2+2x+5 i can't since delta will give me a negative number , so how can i find it ?

also if we take \[\frac{ A }{ x+2 } + \frac{ Bx+c }{ x^2+2x+5 }\]

why do we put Bx+c ? why not just B ? because in the examples that I took before our prof only used B instead of Bx+c ?

Answer 3

I tried to solve it and after multiplying to get 5x^2+6x+2 = A(x^2+2x+5)+(bX+c)(X+2)

if we took x=-2 we will get A=2 and if we take x=0 we will get c=-4 but how about b ?

Answer 4

stop at A(x^2 +2x +5) + (Bx +C) (x+2) = 5x^2 +6x +2
now, if x = -2 you have the second term of the left hand side = 0 , right? and the leftover is A( (-2)^2 +2(-2)+5) = 5(-2)^2 +6(-2)+2
5A = 10
A = 2
get this part?

Answer 5

yeah man already got it a=2 and c=-4 but how about b :D

Answer 6

next, take advantage on the fact that A = 2
let x =0 so, the equation is 2(0^2 +2*0 +5) + (B*0 + C )( 0 +2) = 5*0^2 +6*0 +2
10 + 2C = 2
C = -4
Again, take advantage on the fact that A =2, C =-4
let x = 1 to get B

Answer 7

so b=-5 ?

Answer 8

alright alright , that's good , i'll try to solve it now and get back to you , please stay around :D , also can you please answer my second question " post " ?

Answer 9

:) too busy now, but just post. if it's not me, definitely there is someone else. :)

Answer 10

and yeah , if the roots were define , I should just take the x, but in this case they're not define , so is taking x in this question is up to me ?

Answer 11

oh , thanks man you helped a lot , but if you can just answer these question i'll be so happy :D

Answer 12

I usually pick x as a small number or best way is take the value make one of the term =0 :)

Answer 13

oh , ok , how about my second question , which is about bx+c in the second post ?

Answer 14

because man I got confused , first our prof used only b , now he started using bx+c

Answer 15

that's the rule. if at the denominator, x has degree of 2 ( I mean x^2) the denominator must be in that form

Answer 16

now man i mean about bx+c/ something

Answer 17

why put b(x) and +c ?

Answer 18

because before , our prof used only this form

a/something + b/something

Answer 19

and if we get roots , let's say three , then he puts a/something + b/something + c/something

Answer 20

something is just the roots

Answer 21

in this case if i put b/x^2+2x+5) is it ok ?

Answer 22

that "something " must be a polynomial degree 2. like your problem, the first term,