What is the integral of (1/3)x^3-5x ?

The integral of (1/3)x^3-5x is (x^4}{12}-\frac{5x^2)/2+C

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integral of (1/3)x^3-5x

\int (\frac{1}{3}) x^{3} - 5 x d x

\frac{x ^{4}}{12} - \frac{5 x ^{2}}{2} + C

Solution steps

\int \frac{1}{3} x^{3} - 5 x 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 \frac{1}{3} x^{3} d x - \int 5 x d x

\int \frac{1}{3} x^{3} d x = \frac{x ^{4}}{12}

\int 5 x d x = \frac{5 x ^{2}}{2}

= \frac{x ^{4}}{12} - \frac{5 x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{4}}{12} - \frac{5 x ^{2}}{2} + C

\frac{d}{d x} (9 e^{3 x} x + 6 e^{3 x})

\int (6 x^{2} - 4 x + 5) d x

\int \frac{1}{x ^{3} + 9 x} d x

5 sec (x), at x = \frac{π}{3}

- x^{2} + 3 x, 2 x^{3} - x^{2} - 5 x

What is the integral of (1/3)x^3-5x ?
The integral of (1/3)x^3-5x is (x^4}{12}-\frac{5x^2)/2+C