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

The integral of (3\sqrt[5]{x^2}-sqrt(x)+2) is 15/7 x^{7/5}-2/3 x^{3/2}+2x+C

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

\frac{15}{7} x^{\frac{7}{5}} - \frac{2}{3} x^{\frac{3}{2}} + 2 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 x d x = \frac{2}{3} x^{\frac{3}{2}}

\int 2 d x = 2 x

= \frac{15}{7} x^{\frac{7}{5}} - \frac{2}{3} x^{\frac{3}{2}} + 2 x

Add a constant to the solution

= \frac{15}{7} x^{\frac{7}{5}} - \frac{2}{3} x^{\frac{3}{2}} + 2 x + C

What is the integral of (3\sqrt[5]{x^2}-sqrt(x)+2) ?
The integral of (3\sqrt[5]{x^2}-sqrt(x)+2) is 15/7 x^{7/5}-2/3 x^{3/2}+2x+C