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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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