What is the integral of (4x^{1/3}+1) ?

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

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integral of (4x^{1/3}+1)

\int (4 x^{\frac{1}{3}} + 1) d x

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

Solution steps

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

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

\int 1 d x = x

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

Add a constant to the solution

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

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