What is the integral of (-x^2-3x) ?

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

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integral of (-x^2-3x)

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

- \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} + C

Solution steps

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

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

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

= - \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2}

Add a constant to the solution

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

W (A) = \frac{A}{6}

f (x) = \frac{4 x ^{3}}{( 3 - 2 x ) ^{5}}

x \to 1 lim (x^{2} - 5)

y^{^{'}} = y + x

\frac{72000 + 60 x}{x}

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