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

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

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integral of (6x^2+4x^{1/3})

\int (6 x^{2} + 4 x^{\frac{1}{3}}) d x

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

Solution steps

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

\int 6 x^{2} d x = 2 x^{3}

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

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

Add a constant to the solution

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

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