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

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

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

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

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

Add a constant to the solution

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

\int_{1}^{5} \frac{x ^{2} - 1}{x} d x

\int 9 sin^{2} (θ) d θ

\frac{2 - x ^{2}}{( 2 + x ^{2} ) ^{2}}

\frac{d}{d x} (4 x^{3} ln (x))

f (x) = 110 x - 16 x^{2}, at x = 3.438

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