What is the integral of x(x+4)^{1/3} ?

The integral of x(x+4)^{1/3} is 3(1/7 (x+4)^{7/3}-(x+4)^{4/3})+C

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integral of x(x+4)^{1/3}

\int x (x + 4)^{\frac{1}{3}} d x

3 (\frac{1}{7} (x + 4)^{\frac{7}{3}} - (x + 4)^{\frac{4}{3}}) + C

Solution steps

\int x (x + 4)^{\frac{1}{3}} d x

Apply u-substitution

: \int 3 u^{3} (u^{3} - 4) d u

= \int 3 u^{3} (u^{3} - 4) d u

Take the constant out:

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

= 3 \cdot \int u^{3} (u^{3} - 4) d u

Expand

u^{3} (u^{3} - 4) : u^{6} - 4 u^{3}

= 3 \cdot \int u^{6} - 4 u^{3} 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^{6} d u - \int 4 u^{3} d u)

\int u^{6} d u = \frac{u ^{7}}{7}

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

= 3 (\frac{u ^{7}}{7} - u^{4})

Substitute back

u = (x + 4)^{\frac{1}{3}}

= 3 \frac{( ( x + 4 ) ^{\frac{1}{3}} ) ^{7}}{7} - ((x + 4)^{\frac{1}{3}})^{4}

Simplify

3 \frac{( ( x + 4 ) ^{\frac{1}{3}} ) ^{7}}{7} - ((x + 4)^{\frac{1}{3}})^{4} : 3 (\frac{1}{7} (x + 4)^{\frac{7}{3}} - (x + 4)^{\frac{4}{3}})

= 3 (\frac{1}{7} (x + 4)^{\frac{7}{3}} - (x + 4)^{\frac{4}{3}})

Add a constant to the solution

= 3 (\frac{1}{7} (x + 4)^{\frac{7}{3}} - (x + 4)^{\frac{4}{3}}) + C

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What is the integral of x(x+4)^{1/3} ?
The integral of x(x+4)^{1/3} is 3(1/7 (x+4)^{7/3}-(x+4)^{4/3})+C