What is the integral of (x^2-x-6)/(x^3+x) ?

The integral of (x^2-x-6)/(x^3+x) is 7/2 ln|x^2+1|-arctan(x)-6ln|x|+C

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integral of (x^2-x-6)/(x^3+x)

\int \frac{x ^{2} - x - 6}{x ^{3} + x} d x

\frac{7}{2} ln x^{2} + 1 - arctan (x) - 6 ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{2} - x - 6}{x ^{3} + x} d x

Expand

\frac{x ^{2} - x - 6}{x ^{3} + x} : \frac{x ^{2}}{x ^{3} + x} - \frac{x}{x ^{3} + x} - \frac{6}{x ^{3} + 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{x ^{2}}{x ^{3} + x} d x - \int \frac{x}{x ^{3} + x} d x - \int \frac{6}{x ^{3} + x} d x

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

\int \frac{x}{x ^{3} + x} d x = arctan (x)

\int \frac{6}{x ^{3} + x} d x = 6 (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1)

= \frac{1}{2} ln x^{2} + 1 - arctan (x) - 6 (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1)

Simplify

\frac{1}{2} ln x^{2} + 1 - arctan (x) - 6 (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1) : \frac{7}{2} ln x^{2} + 1 - arctan (x) - 6 ln ∣ x ∣

= \frac{7}{2} ln x^{2} + 1 - arctan (x) - 6 ln ∣ x ∣

Add a constant to the solution

= \frac{7}{2} ln x^{2} + 1 - arctan (x) - 6 ln ∣ x ∣ + C

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What is the integral of (x^2-x-6)/(x^3+x) ?
The integral of (x^2-x-6)/(x^3+x) is 7/2 ln|x^2+1|-arctan(x)-6ln|x|+C