Question

Integrate: (4x^2 + 2x -1) / (x^3 + x^2)

-I know it's an integration using partial fractions. But I seriously CANNOT figure it out. Is this the correct way to split up the fraction? :
A/x + Bx+c/x^2 + D/x+1

And can someone help me out with solving the basic equation? @Mathematics

Answer 1

I Okay, this is my answer. Check it again cuz I'm just done real quick!

(4x^2 + 2x -1) / x^2(x+1)
4x^2 + 2x -1 = A/x + B/x^2 + C/x+2
4x^2 +2x -1 = Ax(x+1) + B(x+1) + Cx^2
4x^2 +2x -1= Ax^2 +Ax +Bx +B + Cx^2
Then we have:
4= A+C
2=A+B
-1= B
=> A=3; B=-1; and C=1
Replace to the original function:
Integration of (4x^2 + 2x -1) / x^3 + x^2 = Interg of 3dx/x - Interg of dx/x^2 + Interg of dx/x=1
= 3ln|x| + 1/x + ln |x+1| + C

Answer 2

just to clarify, when you have (x^2+c)^2 in the denominator, the numerator is one degree less than the degree inside the parenthesis of the denominator. In your middle term, you can consider c=0, and the x^2 should be viewed as (x+0)^2, and the numerator is just a constant (no x term)

Answer 3

Thank you lana! and phi, I'm not sure what you meant.

Answer 4

OHH and I see what I did! I used Bx + C over the x^2 as if it were a quadratic term, where I should have used just B because its a repeated linear term. AGH I have a test in BC next block :(

Answer 5

Just follow my way if u think it's easy and clearly to understand. Hope its helpful and good-luck !

Answer 6

or
=A/x^2 +B/x + C/(x+1)
4x^2+2x-1=A(x+1)+B(x)(x+1) +C(x^2)
when x = 0 A= -1
when x=-1 C= 1
knowing C and A, we can solve for B =3