What is the integral of x\sqrt[3]{6x+1} ?

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

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integral of x\sqrt[3]{6x+1}

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

Substitute back

u = 6 x + 1

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

Simplify

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

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

Add a constant to the solution

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

y = (1 + 2 x)^{2}, (2, 25)

y^{^{''}} + y^{^{'}} - 2 y = 1 - 2 x^{2}

t \to 1 lim (sin (π t))

\frac{d}{d j} (3 i + 2 j)

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