How do you integrate $5/(x^2-1)^2$ using partial fractions?

Question

1806 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-27T01:40:27

$int 5/(x^2-1)^2*dx=5/4*Ln((x+1)/(x-1))-5/2*x/(x^2-1)+C$

$int 5/(x^2-1)^2*dx$

= $int 5/[(x+1)*(x-1)]^2*dx$

I decomposed integrand into basic fractions,

$5/[(x+1)*(x-1)]^2=A/(x+1)+B/(x+1)^2+C/(x-1)+D/(x-1)^2$

After expanding denominator,

$A*(x^3-x^2-x+1)+B*(x^2-2x+1)+C*(x^3+x^2-x-1)+D*(x^2+2x+1)=5$

Set $x=-1$ , so $4B=5$ or $B=5/4$

Set $x=1$ , so $4D=5$ or $D=5/4$

Due to denominator have double roots of $-1$ and $1$ , I took derivative both sides,

$A*(3x^2-2x-1)+B*(2x-2)+C*(3x^2+2x-1)+D*(2x+2)=0$

Set $x=-1$ , so $4A-4B=0$ or $A=B=5/4$

Set $x=1$ , so $4C+4D=0$ or $C=-D=-5/4$

Hence,

$int 5/[(x+1)*(x-1)]^2*dx=5/4*int (dx)/(x+1)+5/4*int (dx)/(x+1)^2-5/4*int (dx)/(x-1)+5/4*int (dx)/(x-1)^2$

= $5/4*Ln(x+1)-5/4*(x+1)^(-1)-5/4*Ln(x-1)-5/4*(x-1)^(-1)+C$

= $5/4*Ln((x+1)/(x-1))-5/2*x/(x^2-1)+C$

How do you integrate 5/(x^2-1)^25/(x^2-1)^2 using partial fractions?