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

The integral of x/(1+x^{1/3)} is 3/5 x^{5/3}-3/4 x^{4/3}+x-3/2 x^{2/3}+3x^{1/3}-3ln|x^{1/3}+1|+C

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

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

\frac{3}{5} x^{\frac{5}{3}} - \frac{3}{4} x^{\frac{4}{3}} + x - \frac{3}{2} x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 3 ln x^{\frac{1}{3}} + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{3 u ^{5}}{1 + u} d u

= \int \frac{3 u ^{5}}{1 + 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 ^{5}}{1 + u} d u

Long division

\frac{u ^{5}}{1 + u} : u^{4} - u^{3} + u^{2} - u + 1 - \frac{1}{u + 1}

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

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

\int u^{4} d u = \frac{u ^{5}}{5}

\int u^{3} d u = \frac{u ^{4}}{4}

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

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

\int 1 d u = u

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

= 3 (\frac{u ^{5}}{5} - \frac{u ^{4}}{4} + \frac{u ^{3}}{3} - \frac{u ^{2}}{2} + u - ln ∣ u + 1 ∣)

Substitute back

u = x^{\frac{1}{3}}

= 3 \frac{( x ^{\frac{1}{3}} ) ^{5}}{5} - \frac{( x ^{\frac{1}{3}} ) ^{4}}{4} + \frac{( x ^{\frac{1}{3}} ) ^{3}}{3} - \frac{( x ^{\frac{1}{3}} ) ^{2}}{2} + x^{\frac{1}{3}} - ln x^{\frac{1}{3}} + 1

Simplify

3 \frac{( x ^{\frac{1}{3}} ) ^{5}}{5} - \frac{( x ^{\frac{1}{3}} ) ^{4}}{4} + \frac{( x ^{\frac{1}{3}} ) ^{3}}{3} - \frac{( x ^{\frac{1}{3}} ) ^{2}}{2} + x^{\frac{1}{3}} - ln x^{\frac{1}{3}} + 1 : \frac{3}{5} x^{\frac{5}{3}} - \frac{3}{4} x^{\frac{4}{3}} + x - \frac{3}{2} x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 3 ln x^{\frac{1}{3}} + 1

= \frac{3}{5} x^{\frac{5}{3}} - \frac{3}{4} x^{\frac{4}{3}} + x - \frac{3}{2} x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 3 ln x^{\frac{1}{3}} + 1

Add a constant to the solution

= \frac{3}{5} x^{\frac{5}{3}} - \frac{3}{4} x^{\frac{4}{3}} + x - \frac{3}{2} x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 3 ln x^{\frac{1}{3}} + 1 + C

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\int \frac{2 x - 1}{x ^{2} - x + 8} d x

\int \frac{x ^{2}}{3 + ( x ^{3} - 4 ) ^{2}} d x

What is the integral of x/(1+x^{1/3)} ?
The integral of x/(1+x^{1/3)} is 3/5 x^{5/3}-3/4 x^{4/3}+x-3/2 x^{2/3}+3x^{1/3}-3ln|x^{1/3}+1|+C