What is the integral of \sqrt[3]{x}-6x ?

The integral of \sqrt[3]{x}-6x is 3/4 x^{4/3}-3x^2+C

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integral of \sqrt[3]{x}-6x

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

Solution steps

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 d x = 3 x^{2}

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

Add a constant to the solution

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

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

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + y})

\frac{s}{4 s ^{2} + 1}

\frac{d}{d x} (\frac{x - 5}{x ^{5}})

6 e^{r}

What is the integral of \sqrt[3]{x}-6x ?
The integral of \sqrt[3]{x}-6x is 3/4 x^{4/3}-3x^2+C