What is the integral of x^3(1/x+6x^2) ?

The integral of x^3(1/x+6x^2) is (x^3)/3+x^6+C

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integral of x^3(1/x+6x^2)

\int x^{3} (\frac{1}{x} + 6 x^{2}) d x

\frac{x ^{3}}{3} + x^{6} + C

Solution steps

\int x^{3} (\frac{1}{x} + 6 x^{2}) d x

Expand

x^{3} (\frac{1}{x} + 6 x^{2}) : x^{2} + 6 x^{5}

= \int x^{2} + 6 x^{5} 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

= \int x^{2} d x + \int 6 x^{5} d x

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

\int 6 x^{5} d x = x^{6}

= \frac{x ^{3}}{3} + x^{6}

Add a constant to the solution

= \frac{x ^{3}}{3} + x^{6} + C

\int 7 x^{6} sin (x^{7}) d x

\int x x^{2} - 3 d x

\frac{1}{s + 2}

8 x + 6, x - x^{2}, - 6, - 1

\int_{- 3}^{3} \frac{14}{x ^{2} - 4 x - 45} d x

What is the integral of x^3(1/x+6x^2) ?
The integral of x^3(1/x+6x^2) is (x^3)/3+x^6+C