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

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integral of x/(2+x^3)

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

Solution steps

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

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

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

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

Take the constant out:

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

Apply u-substitution

: \int \frac{2 ^{\frac{2}{3}} u}{1 + u ^{3}} d u

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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

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

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

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

Substitute back

Simplify

Add a constant to the solution

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