What is the integral of 6/(x+\sqrt[3]{x)} ?

The integral of 6/(x+\sqrt[3]{x)} is 9ln|x^{2/3}+1|+C

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integral of 6/(x+\sqrt[3]{x)}

9 ln x^{\frac{2}{3}} + 1 + C

Solution steps

Take the constant out:

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

Apply u-substitution

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

= 6 \cdot \int \frac{3 u}{u ^{2} + 1} d u

Take the constant out:

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

= 6 \cdot 3 \cdot \int \frac{u}{u ^{2} + 1} d u

Apply u-substitution

: \int \frac{1}{2 v} d v

= 6 \cdot 3 \cdot \int \frac{1}{2 v} d v

Take the constant out:

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

= 6 \cdot 3 \cdot \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 6 \cdot 3 \cdot \frac{1}{2} ln ∣ v ∣

Substitute back

Simplify

= 9 ln x^{\frac{2}{3}} + 1

Add a constant to the solution

= 9 ln x^{\frac{2}{3}} + 1 + C

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What is the integral of 6/(x+\sqrt[3]{x)} ?
The integral of 6/(x+\sqrt[3]{x)} is 9ln|x^{2/3}+1|+C