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

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

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

Take the partial fraction of

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

\int \frac{3 ^{- x}}{x} d x

y (x + 1) + y^{^{'}} = 0, y (- 2) = 1

\frac{\partial}{\partial y} (2 x^{3} y)

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

x = 0, y = 0, y = \frac{- x ^{2} + 8}{9}

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