What is the integral of x^{1/3}+(2x)^{1/3} ?

The integral of x^{1/3}+(2x)^{1/3} is (3*2^{2/3}+6)/(4*2^{2/3)}x^{4/3}+C

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integral of x^{1/3}+(2x)^{1/3}

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

\frac{3 \cdot 2 ^{\frac{2}{3}} + 6}{4 \cdot 2 ^{\frac{2}{3}}} x^{\frac{4}{3}} + C

Solution steps

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= \int x^{\frac{1}{3}} + 2^{\frac{1}{3}} x^{\frac{1}{3}} d x

Apply the Sum Rule:

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

= \int x^{\frac{1}{3}} d x + \int 2^{\frac{1}{3}} x^{\frac{1}{3}} d x

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

\int 2^{\frac{1}{3}} x^{\frac{1}{3}} d x = \frac{3}{2 \cdot 2 ^{\frac{2}{3}}} x^{\frac{4}{3}}

= \frac{3}{4} x^{\frac{4}{3}} + \frac{3}{2 \cdot 2 ^{\frac{2}{3}}} x^{\frac{4}{3}}

Simplify

\frac{3}{4} x^{\frac{4}{3}} + \frac{3}{2 \cdot 2 ^{\frac{2}{3}}} x^{\frac{4}{3}} : \frac{3 \cdot 2 ^{\frac{2}{3}} + 6}{4 \cdot 2 ^{\frac{2}{3}}} x^{\frac{4}{3}}

= \frac{3 \cdot 2 ^{\frac{2}{3}} + 6}{4 \cdot 2 ^{\frac{2}{3}}} x^{\frac{4}{3}}

Add a constant to the solution

= \frac{3 \cdot 2 ^{\frac{2}{3}} + 6}{4 \cdot 2 ^{\frac{2}{3}}} x^{\frac{4}{3}} + C

y^{^{'}} - 2 y = e^{2 t}

y = 13 - x^{2}, y = x^{2} - 5

\frac{d y}{d x} - sin (x) y = 2 sin (x)

\frac{d}{d t} (\frac{d x}{d t}) + \frac{d}{d t} x + 2 x = t^{3}

x^{2} \frac{d y}{d x} = y - x y, y (- 1) = - 3

What is the integral of x^{1/3}+(2x)^{1/3} ?
The integral of x^{1/3}+(2x)^{1/3} is (3*2^{2/3}+6)/(4*2^{2/3)}x^{4/3}+C