What is the integral of 1/(x\sqrt[3]{1+x^5)} ?

The integral of 1/(x\sqrt[3]{1+x^5)} is

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integral of 1/(x\sqrt[3]{1+x^5)}

Solution steps

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 v}{v ^{3} - 1} d v

= \frac{1}{5} \cdot \int \frac{3 v}{v ^{3} - 1} d v

Take the constant out:

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

= \frac{1}{5} \cdot 3 \cdot \int \frac{v}{v ^{3} - 1} d v

Take the partial fraction of

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

= \frac{1}{5} \cdot 3 \cdot \int \frac{1}{3 ( v - 1 )} + \frac{- v + 1}{3 ( v ^{2} + v + 1 )} d v

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

\int \frac{1}{3 ( v - 1 )} d v = \frac{1}{3} ln ∣ v - 1 ∣

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

= \frac{1}{5} \cdot 3 (\frac{1}{3} ln ∣ v - 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 v ^{2} + 4 v}{3} + 23 arctan (\frac{2 v + 1}{3})))

Substitute back

Simplify

Add a constant to the solution

(x^{2} + y) + (x + e^{y}) \frac{d y}{d x} = 0

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

\int sin^{3} (x) d x

x \to - 2 lim (x (x - 1))

\frac{2000 x}{4 x + 75}

What is the integral of 1/(x\sqrt[3]{1+x^5)} ?
The integral of 1/(x\sqrt[3]{1+x^5)} is