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

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Answer 1 · 2017-12-25T14:29:46

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

Reminder

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

Let's perform the decomposition into partial fractions

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

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

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

The denominators are the same, compare the numerators

$1=A(x+1)^2(x-1)^2+B(x)(x-1)^2+C(x)(x-1)^2(x+1)+D(x)(x+1)^2+E(x)(x+1)^2(x-1)$

Let $x=0$ , $=>$ , $1=A$

Let $x=1$ , $=>$ , $1=4D$ , $=>$ , $D=1/4$

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

Coefficients of $x^4$

$0=A+C+E$ , $=>$ , $C+E=-A=-1$

Coefficients of $x^3$

$0=B+D-C+E$ , $=>$ , $B+D=C-E=0$

$C=E=-1/2$

Therefore,

$1/(x(x^2-1)^2)=1/x+(-1/4)/(x+1)^2+(-1/2)/(x+1)+(1/4)/(x-1)^2+(-1/2)/(x-1)$

So,

$int(dx)/(x(x^2-1)^2)=int(dx)/x-int(1/4dx)/(x+1)^2-int(1/2dx)/(x+1)+int(1/4dx)/(x-1)^2-int(1/2dx)/(x-1)$

$=ln(|x|)+1/(4(x+1))-1/2ln(|x+1|)-1/(4(x-1))-1/2ln(|x-1|)+C$

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