What is the integral of (x-1)/(3x^3+6x) ?

The integral of (x-1)/(3x^3+6x) is 1/(3sqrt(2))arctan(x/(sqrt(2)))-1/6 ln|x|+1/12 ln|x^2+2|+C

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integral of (x-1)/(3x^3+6x)

\int \frac{x - 1}{3 x ^{3} + 6 x} d x

\frac{1}{3 2} arctan (\frac{x}{2}) - \frac{1}{6} ln ∣ x ∣ + \frac{1}{12} ln x^{2} + 2 + C

Solution steps

\int \frac{x - 1}{3 x ^{3} + 6 x} d x

Expand

\frac{x - 1}{3 x ^{3} + 6 x} : \frac{x}{3 x ^{3} + 6 x} - \frac{1}{3 x ^{3} + 6 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}{3 x ^{3} + 6 x} d x - \int \frac{1}{3 x ^{3} + 6 x} d x

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

\int \frac{1}{3 x ^{3} + 6 x} d x = \frac{1}{6} ln ∣ x ∣ - \frac{1}{12} ln x^{2} + 2

= \frac{1}{3 2} arctan (\frac{x}{2}) - (\frac{1}{6} ln ∣ x ∣ - \frac{1}{12} ln x^{2} + 2)

Simplify

= \frac{1}{3 2} arctan (\frac{x}{2}) - \frac{1}{6} ln ∣ x ∣ + \frac{1}{12} ln x^{2} + 2

Add a constant to the solution

= \frac{1}{3 2} arctan (\frac{x}{2}) - \frac{1}{6} ln ∣ x ∣ + \frac{1}{12} ln x^{2} + 2 + C

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What is the integral of (x-1)/(3x^3+6x) ?
The integral of (x-1)/(3x^3+6x) is 1/(3sqrt(2))arctan(x/(sqrt(2)))-1/6 ln|x|+1/12 ln|x^2+2|+C