What is the integral of 1/(x^2(x^2-4)^2) ?

The integral of 1/(x^2(x^2-4)^2) is -1/(16x)+3/128 ln|x+2|-1/(64(x+2))-3/128 ln|x-2|-1/(64(x-2))+C

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integral of 1/(x^2(x^2-4)^2)

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

- \frac{1}{16 x} + \frac{3}{128} ln ∣ x + 2 ∣ - \frac{1}{64 ( x + 2 )} - \frac{3}{128} ln ∣ x - 2 ∣ - \frac{1}{64 ( x - 2 )} + C

Solution steps

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

Take the partial fraction of

\frac{1}{x ^{2} ( x ^{2} - 4 ) ^{2}} : \frac{1}{16 x ^{2}} + \frac{3}{128 ( x + 2 )} + \frac{1}{64 ( x + 2 ) ^{2}} - \frac{3}{128 ( x - 2 )} + \frac{1}{64 ( x - 2 ) ^{2}}

= \int \frac{1}{16 x ^{2}} + \frac{3}{128 ( x + 2 )} + \frac{1}{64 ( x + 2 ) ^{2}} - \frac{3}{128 ( x - 2 )} + \frac{1}{64 ( x - 2 ) ^{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}{16 x ^{2}} d x + \int \frac{3}{128 ( x + 2 )} d x + \int \frac{1}{64 ( x + 2 ) ^{2}} d x - \int \frac{3}{128 ( x - 2 )} d x + \int \frac{1}{64 ( x - 2 ) ^{2}} d x

\int \frac{1}{16 x ^{2}} d x = - \frac{1}{16 x}

\int \frac{3}{128 ( x + 2 )} d x = \frac{3}{128} ln ∣ x + 2 ∣

\int \frac{1}{64 ( x + 2 ) ^{2}} d x = - \frac{1}{64 ( x + 2 )}

\int \frac{3}{128 ( x - 2 )} d x = \frac{3}{128} ln ∣ x - 2 ∣

\int \frac{1}{64 ( x - 2 ) ^{2}} d x = - \frac{1}{64 ( x - 2 )}

= - \frac{1}{16 x} + \frac{3}{128} ln ∣ x + 2 ∣ - \frac{1}{64 ( x + 2 )} - \frac{3}{128} ln ∣ x - 2 ∣ - \frac{1}{64 ( x - 2 )}

Add a constant to the solution

= - \frac{1}{16 x} + \frac{3}{128} ln ∣ x + 2 ∣ - \frac{1}{64 ( x + 2 )} - \frac{3}{128} ln ∣ x - 2 ∣ - \frac{1}{64 ( x - 2 )} + C

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What is the integral of 1/(x^2(x^2-4)^2) ?
The integral of 1/(x^2(x^2-4)^2) is -1/(16x)+3/128 ln|x+2|-1/(64(x+2))-3/128 ln|x-2|-1/(64(x-2))+C