What is the integral of 1/((81-x^2)^2) ?

The integral of 1/((81-x^2)^2) is 1/2916 ln|x+9|-1/(324(x+9))-1/2916 ln|x-9|-1/(324(x-9))+C

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integral of 1/((81-x^2)^2)

\int \frac{1}{( 81 - x ^{2} ) ^{2}} d x

\frac{1}{2916} ln ∣ x + 9 ∣ - \frac{1}{324 ( x + 9 )} - \frac{1}{2916} ln ∣ x - 9 ∣ - \frac{1}{324 ( x - 9 )} + C

Solution steps

\int \frac{1}{( 81 - x ^{2} ) ^{2}} d x

Take the partial fraction of

\frac{1}{( 81 - x ^{2} ) ^{2}} : \frac{1}{2916 ( x + 9 )} + \frac{1}{324 ( x + 9 ) ^{2}} - \frac{1}{2916 ( x - 9 )} + \frac{1}{324 ( x - 9 ) ^{2}}

= \int \frac{1}{2916 ( x + 9 )} + \frac{1}{324 ( x + 9 ) ^{2}} - \frac{1}{2916 ( x - 9 )} + \frac{1}{324 ( x - 9 ) ^{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}{2916 ( x + 9 )} d x + \int \frac{1}{324 ( x + 9 ) ^{2}} d x - \int \frac{1}{2916 ( x - 9 )} d x + \int \frac{1}{324 ( x - 9 ) ^{2}} d x

\int \frac{1}{2916 ( x + 9 )} d x = \frac{1}{2916} ln ∣ x + 9 ∣

\int \frac{1}{324 ( x + 9 ) ^{2}} d x = - \frac{1}{324 ( x + 9 )}

\int \frac{1}{2916 ( x - 9 )} d x = \frac{1}{2916} ln ∣ x - 9 ∣

\int \frac{1}{324 ( x - 9 ) ^{2}} d x = - \frac{1}{324 ( x - 9 )}

= \frac{1}{2916} ln ∣ x + 9 ∣ - \frac{1}{324 ( x + 9 )} - \frac{1}{2916} ln ∣ x - 9 ∣ - \frac{1}{324 ( x - 9 )}

Add a constant to the solution

= \frac{1}{2916} ln ∣ x + 9 ∣ - \frac{1}{324 ( x + 9 )} - \frac{1}{2916} ln ∣ x - 9 ∣ - \frac{1}{324 ( x - 9 )} + C

What is the integral of 1/((81-x^2)^2) ?
The integral of 1/((81-x^2)^2) is 1/2916 ln|x+9|-1/(324(x+9))-1/2916 ln|x-9|-1/(324(x-9))+C