What is the integral of 3-7x^2+\sqrt[5]{x} ?

The integral of 3-7x^2+\sqrt[5]{x} is 3x-(7x^3)/3+5/6 x^{6/5}+C

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integral of 3-7x^2+\sqrt[5]{x}

3 x - \frac{7 x ^{3}}{3} + \frac{5}{6} x^{\frac{6}{5}} + 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{5}{6} x^{\frac{6}{5}}

Add a constant to the solution

= 3 x - \frac{7 x ^{3}}{3} + \frac{5}{6} x^{\frac{6}{5}} + C

What is the integral of 3-7x^2+\sqrt[5]{x} ?
The integral of 3-7x^2+\sqrt[5]{x} is 3x-(7x^3)/3+5/6 x^{6/5}+C