How do you use partial fractions to find the integral $int 1/(x^2-1)dx$ ?

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Answer 1 · 2016-12-09T06:28:11

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

We use $a^2-b^2=(a+b)(a-b)$

The denominator is

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

So, the decompostion into partial fractions is

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

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

Therefore,

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

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

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

so,

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

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

$=-1/2ln(∣x+1∣)+1/2ln(∣x-1∣)+C$

How do you use partial fractions to find the integral int 1/(x^2-1)dxint 1/(x^2-1)dx?