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

integral

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

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

\int \frac{5 x ^{4} - 8 x ^{2} + 3 x - 2}{x ^{2}} d x

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

\int \frac{1}{x \cdot ( 1 + x )} d x

4 x^{3}

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