What is the integral of (x^3+3x^2+3x+63)/((x^2-9)^2) ?

The integral of (x^3+3x^2+3x+63)/((x^2-9)^2) is

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integral of (x^3+3x^2+3x+63)/((x^2-9)^2)

\int \frac{x ^{3} + 3 x ^{2} + 3 x + 63}{( x ^{2} - 9 ) ^{2}} d x

\frac{1}{2} ln \frac{x}{3} + 1 + \frac{1}{2} ln \frac{x}{3} - 1 + \frac{1}{3} ln ∣ x + 3 ∣ - \frac{1}{3} ln ∣ x - 3 ∣ - \frac{7}{4 ( x + 3 )} - \frac{13}{4 ( x - 3 )} - \frac{3}{2 ( x ^{2} - 9 )} + C

Solution steps

\int \frac{x ^{3} + 3 x ^{2} + 3 x + 63}{( x ^{2} - 9 ) ^{2}} d x

Expand

\frac{x ^{3} + 3 x ^{2} + 3 x + 63}{( x ^{2} - 9 ) ^{2}} : \frac{x ^{3}}{( x ^{2} - 9 ) ^{2}} + \frac{3 x ^{2}}{( x ^{2} - 9 ) ^{2}} + \frac{3 x}{( x ^{2} - 9 ) ^{2}} + \frac{63}{( x ^{2} - 9 ) ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x ^{3}}{( x ^{2} - 9 ) ^{2}} d x + \int \frac{3 x ^{2}}{( x ^{2} - 9 ) ^{2}} d x + \int \frac{3 x}{( x ^{2} - 9 ) ^{2}} d x + \int \frac{63}{( x ^{2} - 9 ) ^{2}} d x

\int \frac{x ^{3}}{( x ^{2} - 9 ) ^{2}} d x = \frac{1}{2} ln \frac{x}{3} + 1 + \frac{3}{4 ( x + 3 )} + \frac{1}{2} ln \frac{x}{3} - 1 - \frac{3}{4 ( x - 3 )}

\int \frac{3 x ^{2}}{( x ^{2} - 9 ) ^{2}} d x = 3 (- \frac{1}{12} ln ∣ x + 3 ∣ - \frac{1}{4 ( x + 3 )} + \frac{1}{12} ln ∣ x - 3 ∣ - \frac{1}{4 ( x - 3 )})

\int \frac{3 x}{( x ^{2} - 9 ) ^{2}} d x = - \frac{3}{2 ( x ^{2} - 9 )}

\int \frac{63}{( x ^{2} - 9 ) ^{2}} d x = 63 (\frac{1}{108} ln ∣ x + 3 ∣ - \frac{1}{36 ( x + 3 )} - \frac{1}{108} ln ∣ x - 3 ∣ - \frac{1}{36 ( x - 3 )})

= \frac{1}{2} ln \frac{x}{3} + 1 + \frac{3}{4 ( x + 3 )} + \frac{1}{2} ln \frac{x}{3} - 1 - \frac{3}{4 ( x - 3 )} + 3 (- \frac{1}{12} ln ∣ x + 3 ∣ - \frac{1}{4 ( x + 3 )} + \frac{1}{12} ln ∣ x - 3 ∣ - \frac{1}{4 ( x - 3 )}) - \frac{3}{2 ( x ^{2} - 9 )} + 63 (\frac{1}{108} ln ∣ x + 3 ∣ - \frac{1}{36 ( x + 3 )} - \frac{1}{108} ln ∣ x - 3 ∣ - \frac{1}{36 ( x - 3 )})

Simplify

\frac{1}{2} ln \frac{x}{3} + 1 + \frac{3}{4 ( x + 3 )} + \frac{1}{2} ln \frac{x}{3} - 1 - \frac{3}{4 ( x - 3 )} + 3 (- \frac{1}{12} ln ∣ x + 3 ∣ - \frac{1}{4 ( x + 3 )} + \frac{1}{12} ln ∣ x - 3 ∣ - \frac{1}{4 ( x - 3 )}) - \frac{3}{2 ( x ^{2} - 9 )} + 63 (\frac{1}{108} ln ∣ x + 3 ∣ - \frac{1}{36 ( x + 3 )} - \frac{1}{108} ln ∣ x - 3 ∣ - \frac{1}{36 ( x - 3 )}) : \frac{1}{2} ln \frac{x}{3} + 1 + \frac{1}{2} ln \frac{x}{3} - 1 + \frac{1}{3} ln ∣ x + 3 ∣ - \frac{1}{3} ln ∣ x - 3 ∣ - \frac{7}{4 ( x + 3 )} - \frac{13}{4 ( x - 3 )} - \frac{3}{2 ( x ^{2} - 9 )}

= \frac{1}{2} ln \frac{x}{3} + 1 + \frac{1}{2} ln \frac{x}{3} - 1 + \frac{1}{3} ln ∣ x + 3 ∣ - \frac{1}{3} ln ∣ x - 3 ∣ - \frac{7}{4 ( x + 3 )} - \frac{13}{4 ( x - 3 )} - \frac{3}{2 ( x ^{2} - 9 )}

Add a constant to the solution

= \frac{1}{2} ln \frac{x}{3} + 1 + \frac{1}{2} ln \frac{x}{3} - 1 + \frac{1}{3} ln ∣ x + 3 ∣ - \frac{1}{3} ln ∣ x - 3 ∣ - \frac{7}{4 ( x + 3 )} - \frac{13}{4 ( x - 3 )} - \frac{3}{2 ( x ^{2} - 9 )} + C

What is the integral of (x^3+3x^2+3x+63)/((x^2-9)^2) ?
The integral of (x^3+3x^2+3x+63)/((x^2-9)^2) is