Question

How do you find `int ( x + 1)/(x^3 + x^2 -2x) dx` using partial fractions?

Answer 1

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

Explanation:

Perform the decomposition into partial fractions

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

`=(x+1)/(x(x+2)(x-1))`

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

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

The denominators are the same, compare the numerators

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

Let `x=0`, `=>`, `1=-2A`, `=>`, `A=-1/2`

Let `x=-2`, `=>`, `-1=6B`, `=>`, `B=-1/6`

Let `x=1`, `=>`, `2=3C`, `=>`, `C=2/3`

Therefore,

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

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

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