What is the integral of 1/(x^3+3x) ?

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

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integral of 1/(x^3+3x)

\int \frac{1}{x ^{3} + 3 x} d x

\frac{1}{3} ln ∣ x ∣ - \frac{1}{6} ln x^{2} + 3 + C

Solution steps

\int \frac{1}{x ^{3} + 3 x} d x

Take the partial fraction of

\frac{1}{x ^{3} + 3 x} : \frac{1}{3 x} - \frac{x}{3 ( x ^{2} + 3 )}

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

\int \frac{1}{3 x} d x = \frac{1}{3} ln ∣ x ∣

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

= \frac{1}{3} ln ∣ x ∣ - \frac{1}{6} ln x^{2} + 3

Add a constant to the solution

= \frac{1}{3} ln ∣ x ∣ - \frac{1}{6} ln x^{2} + 3 + C

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