What is the integral of x\sqrt[3]{4x+27} ?

The integral of x\sqrt[3]{4x+27} is 1/16 (3/7 (4x+27)^{7/3}-81/4 (4x+27)^{4/3})+C

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integral of x\sqrt[3]{4x+27}

\int x 3 4 x + 27 d x

\frac{1}{16} (\frac{3}{7} (4 x + 27)^{\frac{7}{3}} - \frac{81}{4} (4 x + 27)^{\frac{4}{3}}) + C

Solution steps

\int x 3 4 x + 27 d x

Apply u-substitution

: \int \frac{3 u ( u - 27 )}{16} d u

= \int \frac{3 u ( u - 27 )}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int 3 u (u - 27) d u

Expand

3 u (u - 27) : u^{\frac{4}{3}} - 27 3 u

= \frac{1}{16} \cdot \int u^{\frac{4}{3}} - 27 3 u 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

= \frac{1}{16} (\int u^{\frac{4}{3}} d u - \int 27 3 u d u)

\int u^{\frac{4}{3}} d u = \frac{3}{7} u^{\frac{7}{3}}

\int 27 3 u d u = \frac{81}{4} u^{\frac{4}{3}}

= \frac{1}{16} (\frac{3}{7} u^{\frac{7}{3}} - \frac{81}{4} u^{\frac{4}{3}})

Substitute back

u = 4 x + 27

= \frac{1}{16} (\frac{3}{7} (4 x + 27)^{\frac{7}{3}} - \frac{81}{4} (4 x + 27)^{\frac{4}{3}})

Add a constant to the solution

= \frac{1}{16} (\frac{3}{7} (4 x + 27)^{\frac{7}{3}} - \frac{81}{4} (4 x + 27)^{\frac{4}{3}}) + C

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What is the integral of x\sqrt[3]{4x+27} ?
The integral of x\sqrt[3]{4x+27} is 1/16 (3/7 (4x+27)^{7/3}-81/4 (4x+27)^{4/3})+C