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

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

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

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

\frac{1}{4 x} - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{16} ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int - \frac{1}{4 x ^{2}} - \frac{1}{16 ( x + 2 )} + \frac{1}{16 ( x - 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}{4 x ^{2}} d x - \int \frac{1}{16 ( x + 2 )} d x + \int \frac{1}{16 ( x - 2 )} d x

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

\int \frac{1}{16 ( x + 2 )} d x = \frac{1}{16} ln ∣ x + 2 ∣

\int \frac{1}{16 ( x - 2 )} d x = \frac{1}{16} ln ∣ x - 2 ∣

= - (- \frac{1}{4 x}) - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{16} ln ∣ x - 2 ∣

Simplify

= \frac{1}{4 x} - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{16} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{1}{4 x} - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{16} ln ∣ x - 2 ∣ + C

\frac{d}{d x} (sin (∣ x ∣))

\int \frac{( x ^{4} - 2 x ^{3} + 1 )}{x ^{2}} d x

\frac{d}{d x} (cos^{4} (\frac{x}{4}))

\int (14 x + 1) 7 x^{2} + x d x

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

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