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

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

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

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

\frac{x ^{3}}{3} + \frac{1}{4} ln ∣ x + 2 ∣ + \frac{3}{4} ln ∣ x - 2 ∣ + C

Solution steps

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

Long division

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

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

\int x^{2} d x = \frac{x ^{3}}{3}

\int \frac{x + 1}{x ^{2} - 4} d x = \frac{1}{4} ln ∣ x + 2 ∣ + \frac{3}{4} ln ∣ x - 2 ∣

= \frac{x ^{3}}{3} + \frac{1}{4} ln ∣ x + 2 ∣ + \frac{3}{4} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{1}{4} ln ∣ x + 2 ∣ + \frac{3}{4} ln ∣ x - 2 ∣ + C

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