What is the integral of x\sqrt[3]{2x} ?

The integral of x\sqrt[3]{2x} is \sqrt[3]{2}(1/2 x^2\sqrt[3]{x}-1/14 x^{7/3})+C

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integral of x\sqrt[3]{2x}

\int x 3 2 x d x

32 (\frac{1}{2} x^{2} 3 x - \frac{1}{14} x^{\frac{7}{3}}) + C

Solution steps

\int x 3 2 x d x

3 2 x = 32 3 x

= \int 32 x 3 x d x

Take the constant out:

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

= 32 \cdot \int x 3 x d x

Apply Integration By Parts

: \frac{1}{2} x^{2} 3 x - \int \frac{1}{6} x^{\frac{4}{3}} d x

= 32 (\frac{1}{2} x^{2} 3 x - \int \frac{1}{6} x^{\frac{4}{3}} d x)

\int \frac{1}{6} x^{\frac{4}{3}} d x = \frac{1}{14} x^{\frac{7}{3}}

= 32 (\frac{1}{2} x^{2} 3 x - \frac{1}{14} x^{\frac{7}{3}})

Add a constant to the solution

= 32 (\frac{1}{2} x^{2} 3 x - \frac{1}{14} x^{\frac{7}{3}}) + C

f (t) = \frac{5 t ^{2} + t + 6}{t}

\int \frac{7 x}{( 1 - x ^{2} ) ^{3}} d x

x \to π lim (\frac{cos ( x )}{x})

\frac{d}{d x} (\frac{2 x - 3}{x ^{2} + 1})

y^{^{'}} = 2 x y + 3 y - 4 x - 6

What is the integral of x\sqrt[3]{2x} ?
The integral of x\sqrt[3]{2x} is \sqrt[3]{2}(1/2 x^2\sqrt[3]{x}-1/14 x^{7/3})+C