How do you evaluate the integral $int (x+2)/(x^2-2x-3)$ ?

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Answer 1 · 2017-02-01T14:31:15

The answer is $=-1/4ln(|x+1|)+5/4ln(|x-3|)+C$

Let's factorise the denominator

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

Let's perform the decomposition into partial fractions

$(x+2)/((x+1)(x-3))=A/(x+1)+B/(x-3)$

$=(A(x-3)+B(x+1))/((x+1)(x-3))$

The denominators are the same, we can equalize the numerators

$(x+2)=A(x-3)+B(x+1)$

Let $x=-1$ , $=>$ , $1=-4A$ , $=>$ , $A=-1/4$

Let $x=3$ , $=>$ , $5=4B$ , $B=5/4$

Therefore,

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

$int((x+2)dx)/((x+1)(x-3))=-1/4intdx/(x+1)+5/4intdx/(x-3)$

$=-1/4ln(|x+1|)+5/4ln(|x-3|)+C$

How do you evaluate the integral int (x+2)/(x^2-2x-3)int (x+2)/(x^2-2x-3)?