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

The integral of 7x(x^2+\sqrt[3]{x}) is (7x^4)/4+3x^{7/3}+C

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

\frac{7 x ^{4}}{4} + 3 x^{\frac{7}{3}} + C

Solution steps

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

Expand

= 7 \cdot \int x^{3} + x^{\frac{4}{3}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 7 (\int x^{3} d x + \int x^{\frac{4}{3}} d x)

\int x^{3} d x = \frac{x ^{4}}{4}

\int x^{\frac{4}{3}} d x = \frac{3}{7} x^{\frac{7}{3}}

= 7 (\frac{x ^{4}}{4} + \frac{3}{7} x^{\frac{7}{3}})

Simplify

7 (\frac{x ^{4}}{4} + \frac{3}{7} x^{\frac{7}{3}}) : \frac{7 x ^{4}}{4} + 3 x^{\frac{7}{3}}

= \frac{7 x ^{4}}{4} + 3 x^{\frac{7}{3}}

Add a constant to the solution

= \frac{7 x ^{4}}{4} + 3 x^{\frac{7}{3}} + C

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What is the integral of 7x(x^2+\sqrt[3]{x}) ?
The integral of 7x(x^2+\sqrt[3]{x}) is (7x^4)/4+3x^{7/3}+C