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

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Answer 1 · 2017-08-02T17:03:58

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

Let's simplify the quotient

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

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

Now, we perform the decomposition into partial fractions

$(x+2)/(x^2-1)=A/(x+1)+B/(x-1)=(A(x-1)+B(x+1))/((x-1)(x+1))$

The numerators are the same. we compare the numerators

$x+2=A(x-1)+B(x+1))$

Let $x=1$ , $=>$ , $3=2B$ , $=>$ , $B=3/2$

Let $x=-1$ , $=>$ , $1=-2A$ , $=>$ , $A=-1/2$

Therefore,

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

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

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

How do you integrate int (x^2+x+1)/(1-x^2)int (x^2+x+1)/(1-x^2) using partial fractions?