How do you use partial fractions to find the integral $int (6x^2+1)/(x^2(x-1)^3)dx$ ?

Question

1497 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-12T17:00:50

$-3lnx+1/x-3ln(x-1)+2/(x-1)-7/2*(x-1)^(-2)+C$

I decomposed integrand into basic fractions,

$(6x^2+1)/[x^2*(x-1)^2]$

= $-3/x-1/x^2+3/(x-1)-2/(x-1)^2+7/(x-1)^3$

Hence,

$int ((6x^2+1)dx)/[x^2*(x-1)^3]$

= $-int (3dx)/x$ - $int dx/x^2$ - $int (3dx)/(x-1)$ - $int (2dx)/(x-1)^2$ + $int (7dx)/(x-1)^3$

= $-3lnx+1/x-3ln(x-1)+2/(x-1)-7/2*(x-1)^(-2)+C$

How do you use partial fractions to find the integral int (6x^2+1)/(x^2(x-1)^3)dxint (6x^2+1)/(x^2(x-1)^3)dx?