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

The integral of (x^2)/((x^2-4)(x^2+1)) is -1/5 ln|x+2|+1/5 ln|x-2|+1/5 arctan(x)+C

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

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

- \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 2 ∣ + \frac{1}{5} arctan (x) + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2}}{( x ^{2} - 4 ) ( x ^{2} + 1 )} : - \frac{1}{5 ( x + 2 )} + \frac{1}{5 ( x - 2 )} + \frac{1}{5 ( x ^{2} + 1 )}

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

\int \frac{1}{5 ( x + 2 )} d x = \frac{1}{5} ln ∣ x + 2 ∣

\int \frac{1}{5 ( x - 2 )} d x = \frac{1}{5} ln ∣ x - 2 ∣

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

= - \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 2 ∣ + \frac{1}{5} arctan (x)

Add a constant to the solution

= - \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 2 ∣ + \frac{1}{5} arctan (x) + C

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What is the integral of (x^2)/((x^2-4)(x^2+1)) ?
The integral of (x^2)/((x^2-4)(x^2+1)) is -1/5 ln|x+2|+1/5 ln|x-2|+1/5 arctan(x)+C