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

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Answer 1 · 2018-03-13T10:17:37

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

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,

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

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

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

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