What is the integral of 2x^2+\sqrt[3]{x}-1 ?

The integral of 2x^2+\sqrt[3]{x}-1 is (2x^3)/3+3/4 x^{4/3}-x+C

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integral of 2x^2+\sqrt[3]{x}-1

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

\int 1 d x = x

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

Add a constant to the solution

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

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What is the integral of 2x^2+\sqrt[3]{x}-1 ?
The integral of 2x^2+\sqrt[3]{x}-1 is (2x^3)/3+3/4 x^{4/3}-x+C