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

The integral of (3x^{1/3})/(1+x^{2/3)} is 9/2 (1+x^{2/3}-ln|1+x^{2/3}|)+C

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

= 3 \cdot 3 \cdot \frac{1}{2} \cdot \int 1 - \frac{1}{v} 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

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

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= 3 \cdot 3 \cdot \frac{1}{2} (v - ln ∣ v ∣)

Substitute back

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

Simplify

3 \cdot 3 \cdot \frac{1}{2} (1 + (x^{\frac{1}{3}})^{2} - ln 1 + (x^{\frac{1}{3}})^{2}) : \frac{9}{2} (1 + x^{\frac{2}{3}} - ln 1 + x^{\frac{2}{3}})

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

Add a constant to the solution

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

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What is the integral of (3x^{1/3})/(1+x^{2/3)} ?
The integral of (3x^{1/3})/(1+x^{2/3)} is 9/2 (1+x^{2/3}-ln|1+x^{2/3}|)+C