What is the integral of x^{1/3}+6 ?

The integral of x^{1/3}+6 is 3/4 x^{4/3}+6x+C

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integral of x^{1/3}+6

\int x^{\frac{1}{3}} + 6 d x

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

Solution steps

\int x^{\frac{1}{3}} + 6 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 6 d x

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

\int 6 d x = 6 x

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

Add a constant to the solution

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

\frac{d}{d x} (- e^{- 3 x})

e^{2 t}

\frac{s + 1}{0.1 s + 1}

\frac{d}{d x} (\frac{sin ( 4 x )}{x})

\int_{1}^{2} 12 x^{2} 2^{x^{3}} d x

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