What is the integral of 1/((100-x^2)^2) ?

The integral of 1/((100-x^2)^2) is 1/4000 ln|x+10|-1/(400(x+10))-1/4000 ln|x-10|-1/(400(x-10))+C

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integral of 1/((100-x^2)^2)

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

\frac{1}{4000} ln ∣ x + 10 ∣ - \frac{1}{400 ( x + 10 )} - \frac{1}{4000} ln ∣ x - 10 ∣ - \frac{1}{400 ( x - 10 )} + C

Solution steps

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

Take the partial fraction of

\frac{1}{( 100 - x ^{2} ) ^{2}} : \frac{1}{4000 ( x + 10 )} + \frac{1}{400 ( x + 10 ) ^{2}} - \frac{1}{4000 ( x - 10 )} + \frac{1}{400 ( x - 10 ) ^{2}}

= \int \frac{1}{4000 ( x + 10 )} + \frac{1}{400 ( x + 10 ) ^{2}} - \frac{1}{4000 ( x - 10 )} + \frac{1}{400 ( x - 10 ) ^{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}{4000 ( x + 10 )} d x + \int \frac{1}{400 ( x + 10 ) ^{2}} d x - \int \frac{1}{4000 ( x - 10 )} d x + \int \frac{1}{400 ( x - 10 ) ^{2}} d x

\int \frac{1}{4000 ( x + 10 )} d x = \frac{1}{4000} ln ∣ x + 10 ∣

\int \frac{1}{400 ( x + 10 ) ^{2}} d x = - \frac{1}{400 ( x + 10 )}

\int \frac{1}{4000 ( x - 10 )} d x = \frac{1}{4000} ln ∣ x - 10 ∣

\int \frac{1}{400 ( x - 10 ) ^{2}} d x = - \frac{1}{400 ( x - 10 )}

= \frac{1}{4000} ln ∣ x + 10 ∣ - \frac{1}{400 ( x + 10 )} - \frac{1}{4000} ln ∣ x - 10 ∣ - \frac{1}{400 ( x - 10 )}

Add a constant to the solution

= \frac{1}{4000} ln ∣ x + 10 ∣ - \frac{1}{400 ( x + 10 )} - \frac{1}{4000} ln ∣ x - 10 ∣ - \frac{1}{400 ( x - 10 )} + C

(- 2.3) (6.2)

\int \frac{6 x}{6 + x} d x

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

\frac{\partial}{\partial x} (\frac{2 z}{x + y})

\frac{d}{d x} (\frac{2 x + 4 x ^{2}}{( 1 + 4 x ) ^{2}})

What is the integral of 1/((100-x^2)^2) ?
The integral of 1/((100-x^2)^2) is 1/4000 ln|x+10|-1/(400(x+10))-1/4000 ln|x-10|-1/(400(x-10))+C