Question 564861
First decompose the given fraction {{{(x^2+1)/(x^3-x^2-6x)}}} into its partial fractions:
Factor the denominator:
{{{(x^2+1)/(x)(x+2)(x-3)}}} now we set:
{{{(x^2+1)/(x^3-x^2-6x) = A/x+B/(x+2)+C/(x-3)}}} Now add the fractions on the right side.
{{{(x^2+1)/(x^3-x^2-6x) = (A(x+2)(x-3)+B(x)(x-3)+C(x)(x+2))/(x^3-x^2-6x)}}} Now since the denominators are equal, the numerators must be equal, so we can set:
{{{x^2+1 = A(x+2)(x-3)+B(x)(x-3)+C(x)(x+2)}}} This must be true for all x so it is true for the x-values that make:
{{{x = 0}}}
{{{x+2 = 0}}} so {{{x = -2}}} and 
{{{x-3 = 0}}} so {{{x = 3}}} 
So we can solve for A, B, and C by letting x = 0, then x = -2, and x = 3.
After some standard algebra, you'll find that:
{{{A = -1/6}}}, {{{B = -1/2}}}, and {{{C = 2/3}}} we can now write the partial fraction:
{{{(x^2+1)/(x^3-x^2-6x) = -1/6x-1/2(x+2)+2/3(x-3)}}}
Now when you integrate these partial fractions you'll get the answers:
{{{(-1/6)lnx-(1/2)ln(abs(x+2))+(2/3)ln(abs(x-3))+C}}}