What is the integral of 5x\sqrt[3]{(6x-1)^5} ?

The integral of 5x\sqrt[3]{(6x-1)^5} is 5/352 (16x+1)(6x-1)^{8/3}+C

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integral of 5x\sqrt[3]{(6x-1)^5}

\frac{5}{352} (16 x + 1) (6 x - 1)^{\frac{8}{3}} + 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{u ^{\frac{5}{3}} ( u + 1 )}{36} d u

= 5 \cdot \int \frac{u ^{\frac{5}{3}} ( u + 1 )}{36} d u

Take the constant out:

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

= 5 \cdot \frac{1}{36} \cdot \int u^{\frac{5}{3}} (u + 1) d u

Expand

u^{\frac{5}{3}} (u + 1) : u^{\frac{8}{3}} + u^{\frac{5}{3}}

= 5 \cdot \frac{1}{36} \cdot \int u^{\frac{8}{3}} + u^{\frac{5}{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

= 5 \cdot \frac{1}{36} (\int u^{\frac{8}{3}} d u + \int u^{\frac{5}{3}} d u)

\int u^{\frac{8}{3}} d u = \frac{3}{11} u^{\frac{11}{3}}

\int u^{\frac{5}{3}} d u = \frac{3}{8} u^{\frac{8}{3}}

= 5 \cdot \frac{1}{36} (\frac{3}{11} u^{\frac{11}{3}} + \frac{3}{8} u^{\frac{8}{3}})

Substitute back

u = 6 x - 1

= 5 \cdot \frac{1}{36} (\frac{3}{11} (6 x - 1)^{\frac{11}{3}} + \frac{3}{8} (6 x - 1)^{\frac{8}{3}})

Simplify

5 \cdot \frac{1}{36} (\frac{3}{11} (6 x - 1)^{\frac{11}{3}} + \frac{3}{8} (6 x - 1)^{\frac{8}{3}}) : \frac{5}{352} (16 x + 1) (6 x - 1)^{\frac{8}{3}}

= \frac{5}{352} (16 x + 1) (6 x - 1)^{\frac{8}{3}}

Add a constant to the solution

= \frac{5}{352} (16 x + 1) (6 x - 1)^{\frac{8}{3}} + C

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What is the integral of 5x\sqrt[3]{(6x-1)^5} ?
The integral of 5x\sqrt[3]{(6x-1)^5} is 5/352 (16x+1)(6x-1)^{8/3}+C