What is the integral of 1/((x^2-36)^2) ?

The integral of 1/((x^2-36)^2) is 1/864 ln|x+6|-1/(144(x+6))-1/864 ln|x-6|-1/(144(x-6))+C

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integral of 1/((x^2-36)^2)

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

\frac{1}{864} ln ∣ x + 6 ∣ - \frac{1}{144 ( x + 6 )} - \frac{1}{864} ln ∣ x - 6 ∣ - \frac{1}{144 ( x - 6 )} + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x ^{2} - 36 ) ^{2}} : \frac{1}{864 ( x + 6 )} + \frac{1}{144 ( x + 6 ) ^{2}} - \frac{1}{864 ( x - 6 )} + \frac{1}{144 ( x - 6 ) ^{2}}

= \int \frac{1}{864 ( x + 6 )} + \frac{1}{144 ( x + 6 ) ^{2}} - \frac{1}{864 ( x - 6 )} + \frac{1}{144 ( x - 6 ) ^{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}{864 ( x + 6 )} d x + \int \frac{1}{144 ( x + 6 ) ^{2}} d x - \int \frac{1}{864 ( x - 6 )} d x + \int \frac{1}{144 ( x - 6 ) ^{2}} d x

\int \frac{1}{864 ( x + 6 )} d x = \frac{1}{864} ln ∣ x + 6 ∣

\int \frac{1}{144 ( x + 6 ) ^{2}} d x = - \frac{1}{144 ( x + 6 )}

\int \frac{1}{864 ( x - 6 )} d x = \frac{1}{864} ln ∣ x - 6 ∣

\int \frac{1}{144 ( x - 6 ) ^{2}} d x = - \frac{1}{144 ( x - 6 )}

= \frac{1}{864} ln ∣ x + 6 ∣ - \frac{1}{144 ( x + 6 )} - \frac{1}{864} ln ∣ x - 6 ∣ - \frac{1}{144 ( x - 6 )}

Add a constant to the solution

= \frac{1}{864} ln ∣ x + 6 ∣ - \frac{1}{144 ( x + 6 )} - \frac{1}{864} ln ∣ x - 6 ∣ - \frac{1}{144 ( x - 6 )} + C

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What is the integral of 1/((x^2-36)^2) ?
The integral of 1/((x^2-36)^2) is 1/864 ln|x+6|-1/(144(x+6))-1/864 ln|x-6|-1/(144(x-6))+C