What is the integral of x\sqrt[3]{6x+7} ?

The integral of x\sqrt[3]{6x+7} is 1/336 (4(6x+7)^{7/3}-49(6x+7)^{4/3})+C

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integral of x\sqrt[3]{6x+7}

\frac{1}{336} (4 (6 x + 7)^{\frac{7}{3}} - 49 (6 x + 7)^{\frac{4}{3}}) + C

Solution steps

Apply u-substitution

Take the constant out:

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

Expand

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

= \frac{1}{36} (\frac{3}{7} u^{\frac{7}{3}} - \frac{21}{4} u^{\frac{4}{3}})

Substitute back

u = 6 x + 7

= \frac{1}{36} (\frac{3}{7} (6 x + 7)^{\frac{7}{3}} - \frac{21}{4} (6 x + 7)^{\frac{4}{3}})

Simplify

\frac{1}{36} (\frac{3}{7} (6 x + 7)^{\frac{7}{3}} - \frac{21}{4} (6 x + 7)^{\frac{4}{3}}) : \frac{1}{336} (4 (6 x + 7)^{\frac{7}{3}} - 49 (6 x + 7)^{\frac{4}{3}})

= \frac{1}{336} (4 (6 x + 7)^{\frac{7}{3}} - 49 (6 x + 7)^{\frac{4}{3}})

Add a constant to the solution

= \frac{1}{336} (4 (6 x + 7)^{\frac{7}{3}} - 49 (6 x + 7)^{\frac{4}{3}}) + C

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 5 y = 2 e^{- 2 x}

\frac{d y}{d t} = 4 y + y^{4}, y (0) = 1

\frac{d}{d x} (\frac{- 24}{x ^{4}})

\int 2 + y e^{y} d y

x \to 0 - lim (\frac{x}{sin ( 3 x )})

What is the integral of x\sqrt[3]{6x+7} ?
The integral of x\sqrt[3]{6x+7} is 1/336 (4(6x+7)^{7/3}-49(6x+7)^{4/3})+C