What is the integral of 3-7x^2+\sqrt[7]{x} ?

The integral of 3-7x^2+\sqrt[7]{x} is 3x-(7x^3)/3+7/8 x^{8/7}+C

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integral of 3-7x^2+\sqrt[7]{x}

3 x - \frac{7 x ^{3}}{3} + \frac{7}{8} x^{\frac{8}{7}} + 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 3 d x = 3 x

\int 7 x^{2} d x = \frac{7 x ^{3}}{3}

= 3 x - \frac{7 x ^{3}}{3} + \frac{7}{8} x^{\frac{8}{7}}

Add a constant to the solution

= 3 x - \frac{7 x ^{3}}{3} + \frac{7}{8} x^{\frac{8}{7}} + C

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What is the integral of 3-7x^2+\sqrt[7]{x} ?
The integral of 3-7x^2+\sqrt[7]{x} is 3x-(7x^3)/3+7/8 x^{8/7}+C