What is the integral of 1/((x^2-9)^2) ?

The integral of 1/((x^2-9)^2) is 1/108 ln|x+3|-1/(36(x+3))-1/108 ln|x-3|-1/(36(x-3))+C

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integral of 1/((x^2-9)^2)

\int \frac{1}{( x ^{2} - 9 ) ^{2}} d x

\frac{1}{108} ln ∣ x + 3 ∣ - \frac{1}{36 ( x + 3 )} - \frac{1}{108} ln ∣ x - 3 ∣ - \frac{1}{36 ( x - 3 )} + C

Solution steps

\int \frac{1}{( x ^{2} - 9 ) ^{2}} d x

Take the partial fraction of

\frac{1}{( x ^{2} - 9 ) ^{2}} : \frac{1}{108 ( x + 3 )} + \frac{1}{36 ( x + 3 ) ^{2}} - \frac{1}{108 ( x - 3 )} + \frac{1}{36 ( x - 3 ) ^{2}}

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

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{1}{108 ( x + 3 )} d x + \int \frac{1}{36 ( x + 3 ) ^{2}} d x - \int \frac{1}{108 ( x - 3 )} d x + \int \frac{1}{36 ( x - 3 ) ^{2}} d x

\int \frac{1}{108 ( x + 3 )} d x = \frac{1}{108} ln ∣ x + 3 ∣

\int \frac{1}{36 ( x + 3 ) ^{2}} d x = - \frac{1}{36 ( x + 3 )}

\int \frac{1}{108 ( x - 3 )} d x = \frac{1}{108} ln ∣ x - 3 ∣

\int \frac{1}{36 ( x - 3 ) ^{2}} d x = - \frac{1}{36 ( x - 3 )}

= \frac{1}{108} ln ∣ x + 3 ∣ - \frac{1}{36 ( x + 3 )} - \frac{1}{108} ln ∣ x - 3 ∣ - \frac{1}{36 ( x - 3 )}

Add a constant to the solution

= \frac{1}{108} ln ∣ x + 3 ∣ - \frac{1}{36 ( x + 3 )} - \frac{1}{108} ln ∣ x - 3 ∣ - \frac{1}{36 ( x - 3 )} + C

\frac{d}{d x} (3 x - 8)

f (x) = (1 + 3 x)^{10}, at x = 0

x \to 2 lim (x^{2} + 4 x)

2 y \frac{d y}{d x} + 2 = y^{2} + 2 x

\frac{d}{d x} ((2 x - 2)^{3} (3 x - 1)^{2})

What is the integral of 1/((x^2-9)^2) ?
The integral of 1/((x^2-9)^2) is 1/108 ln|x+3|-1/(36(x+3))-1/108 ln|x-3|-1/(36(x-3))+C