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

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

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integral of 3/(1+x^3)

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

ln ∣ x + 1 ∣ + \frac{1}{2} (- ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x - 1}{3})) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

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

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

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

\int \frac{- x + 2}{3 ( x ^{2} - x + 1 )} d x = \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3}))

= 3 (\frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3})))

Simplify

3 (\frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3}))) : ln ∣ x + 1 ∣ + \frac{1}{2} (- ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x - 1}{3}))

= ln ∣ x + 1 ∣ + \frac{1}{2} (- ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x - 1}{3}))

Add a constant to the solution

= ln ∣ x + 1 ∣ + \frac{1}{2} (- ln \frac{4 x ^{2} - 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x - 1}{3})) + C

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