Question

Question #99cca

Answer 1

`1/2x^2-2x-1/2ln|x+1|+1/6ln|x-1|+16/3ln|x+2|+C`

Explanation:

`intx^4/(x^3+2x^2-x-2)dx`

`=int(x^4+2x^3-x^2-2x)/(x^3+2x^2-x-2)+(-2x^3+x^2+2x)/(x^3+2x^2-x-2)dx` (split fraction so the first fraction will be divisible by the denominator. this reduces the power of the numerator of the resulting fraction from 4 to 3)

`=intx+(-2x^3-4x^2+2x+4)/(x^3+2x^2-x-2)+(5x^2-4)/(x^3+2x^2-x-2)dx` (split fraction again)

`=1/2x^2+int-2+(5x^2-4)/((x^2-1)(x+2))dx` (power rule integration and factorize denominator)

`=1/2x^2-2x+int(5x^2-4)/((x+1)(x-1)(x+2))dx`

now use partial fraction decomposition for the fraction:

`(5x^2-4)/((x+1)(x-1)(x+2))=A/(x+1)+B/(x-1)+C/(x+2)` (A, B, and C are constants)

`5x^2-4=A(x-1)(x+2)+B(x+1)(x+2)+C(x+1)(x-1)`

if x=-1:

`5(-1)^2-4=A(-1-1)(-1+2)+B(-1+1)(-1+2)+C(-1+1)(-1-1)`
`1=A(-2)(1)+B(0)(1)+C(0)(-2)`
`1=A(-2)`
`-1/2=A`

if x=1:

`5(1)^2-4=A(1-1)(1+2)+B(1+1)(1+2)+C(1+1)(1-1)`
`1=A(0)(3)+B(2)(3)+C(2)(0)`
`1=B(6)`
`1/6=B`

if x=-2:

`5(-2)^2-4=A(-2-1)(-2+2)+B(-2+1)(-2+2)+C(-2+1)(-2-1)`
`16=A(-3)(0)+B(-1)(0)+C(-1)(-3)`
`16=C(3)`
`16/3=C`

substitute:
`=1/2x^2-2x+int(-1/2)/(x+1)+(1/6)/(x-1)+(16/3)/(x+2)dx`
`=1/2x^2-2x-1/2ln|x+1|+1/6ln|x-1|+16/3ln|x+2|+C` (integration to natural log)

Answer 2

The answer is `=x^2/2-2x+16/3ln(|x+2|)-1/2ln(|x+1|)+1/6ln(|x-1|)+C`

Explanation:

The degree of the numerator is greater than the degree of the denominator, perform a polynomial long division.

`x^4/(x^3+2x^2-x-2)=(x-2)+((5x^2-4))/((x^3+2x^2-x-2))`

Factorise the denominator,

`x^3+2x^2-x-2=x^2(x+2)-(x+2)=(x+2)(x^2-1)=(x+2)(x+1)(x-1)`

`x^4/(x^3+2x^2-x-2)=(x-2)+((5x^2-4))/((x+2)(x+1)(x-1))`

Perform a decompostion into partial fractions

`((5x^2-4))/((x+2)(x+1)(x-1))=A/(x+2)+B/(x+1)+C/(x-1)`

`=(A(x+1)(x-1)+B(x+2)(x-1)+C(x+2)(x+1))/((x+2)(x+1)(x-1))`

The denominator is the same, compare the numerator

`5x^2-4=A(x+1)(x-1)+B(x+2)(x-1)+C(x+2)(x+1)`

Let `x=-2`, `=>`, `16=3A`, `=>`, `A=16/3`

Let `x=-1`, `=>`, `1=-2B`, `=>`, `B=-1/2`

Let `x=1`, `=>`, `1=6C`, `=>`, `C=1/6`

Therefore,

`x^4/(x^3+2x^2-x-2)=(x-2)+(16/3)/(x+2)-(1/2)/(x+1)+(1/6)/(x-1)`

Perform the integration

`int(x^4dx)/(x^3+2x^2-x-2)=int(x-2)dx+int(16/3dx)/(x+2)-int(1/2dx)/(x+1)+int(1/6dx)/(x-1)`

`=x^2/2-2x+16/3ln(|x+2|)-1/2ln(|x+1|)+1/6ln(|x-1|)+C`