What is the integral of 1/((25-x^2)^2) ?

The integral of 1/((25-x^2)^2) is 1/500 ln|x+5|-1/(100(x+5))-1/500 ln|x-5|-1/(100(x-5))+C

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integral of 1/((25-x^2)^2)

\int \frac{1}{( 25 - x ^{2} ) ^{2}} d x

\frac{1}{500} ln ∣ x + 5 ∣ - \frac{1}{100 ( x + 5 )} - \frac{1}{500} ln ∣ x - 5 ∣ - \frac{1}{100 ( x - 5 )} + C

Solution steps

\int \frac{1}{( 25 - x ^{2} ) ^{2}} d x

Take the partial fraction of

\frac{1}{( 25 - x ^{2} ) ^{2}} : \frac{1}{500 ( x + 5 )} + \frac{1}{100 ( x + 5 ) ^{2}} - \frac{1}{500 ( x - 5 )} + \frac{1}{100 ( x - 5 ) ^{2}}

= \int \frac{1}{500 ( x + 5 )} + \frac{1}{100 ( x + 5 ) ^{2}} - \frac{1}{500 ( x - 5 )} + \frac{1}{100 ( x - 5 ) ^{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}{500 ( x + 5 )} d x + \int \frac{1}{100 ( x + 5 ) ^{2}} d x - \int \frac{1}{500 ( x - 5 )} d x + \int \frac{1}{100 ( x - 5 ) ^{2}} d x

\int \frac{1}{500 ( x + 5 )} d x = \frac{1}{500} ln ∣ x + 5 ∣

\int \frac{1}{100 ( x + 5 ) ^{2}} d x = - \frac{1}{100 ( x + 5 )}

\int \frac{1}{500 ( x - 5 )} d x = \frac{1}{500} ln ∣ x - 5 ∣

\int \frac{1}{100 ( x - 5 ) ^{2}} d x = - \frac{1}{100 ( x - 5 )}

= \frac{1}{500} ln ∣ x + 5 ∣ - \frac{1}{100 ( x + 5 )} - \frac{1}{500} ln ∣ x - 5 ∣ - \frac{1}{100 ( x - 5 )}

Add a constant to the solution

= \frac{1}{500} ln ∣ x + 5 ∣ - \frac{1}{100 ( x + 5 )} - \frac{1}{500} ln ∣ x - 5 ∣ - \frac{1}{100 ( x - 5 )} + C

What is the integral of 1/((25-x^2)^2) ?
The integral of 1/((25-x^2)^2) is 1/500 ln|x+5|-1/(100(x+5))-1/500 ln|x-5|-1/(100(x-5))+C