What is the integral of 1/(x+\sqrt[3]{x-2)} ?

The integral of 1/(x+\sqrt[3]{x-2)} is

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integral of 1/(x+\sqrt[3]{x-2)}

Solution steps

Apply u-substitution

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

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

Take the constant out:

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

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

Take the partial fraction of

\frac{u ^{2}}{u ^{3} + 2 + u} : \frac{1}{4 ( u + 1 )} + \frac{3 u - 2}{4 ( u ^{2} - u + 2 )}

= 3 \cdot \int \frac{1}{4 ( u + 1 )} + \frac{3 u - 2}{4 ( u ^{2} - u + 2 )} 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

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

\int \frac{1}{4 ( u + 1 )} d u = \frac{1}{4} ln ∣ u + 1 ∣

\int \frac{3 u - 2}{4 ( u ^{2} - u + 2 )} d u = \frac{1}{8 7} (37 ln 4 u^{2} - 4 u + 8 - 2 arctan (\frac{2 u - 1}{7}))

= 3 (\frac{1}{4} ln ∣ u + 1 ∣ + \frac{1}{8 7} (37 ln 4 u^{2} - 4 u + 8 - 2 arctan (\frac{2 u - 1}{7})))

Substitute back

Simplify

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What is the integral of 1/(x+\sqrt[3]{x-2)} ?
The integral of 1/(x+\sqrt[3]{x-2)} is