What is the integral of x/(2x^3+1) ?

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integral of x/(2x^3+1)

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

Solution steps

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

Apply u-substitution

Take the constant out:

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

Apply u-substitution

: \int \frac{3 w}{w ^{3} + 1} d w

= \frac{1}{3 \cdot 2 ^{\frac{2}{3}}} \cdot \int \frac{3 w}{w ^{3} + 1} d w

Take the constant out:

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

= \frac{1}{3 \cdot 2 ^{\frac{2}{3}}} \cdot 3 \cdot \int \frac{w}{w ^{3} + 1} d w

Take the partial fraction of

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Substitute back

Simplify

Add a constant to the solution

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