What is the integral of (-x^2+6x-5) ?

The integral of (-x^2+6x-5) is -(x^3)/3+3x^2-5x+C

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integral of (-x^2+6x-5)

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

- \frac{x ^{3}}{3} + 3 x^{2} - 5 x + C

Solution steps

\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 d x - \int 5 d x

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

\int 6 x d x = 3 x^{2}

\int 5 d x = 5 x

= - \frac{x ^{3}}{3} + 3 x^{2} - 5 x

Add a constant to the solution

= - \frac{x ^{3}}{3} + 3 x^{2} - 5 x + C

\int \frac{x - 2}{3 x + 6} d x

\frac{d}{d x} (16 - x^{2} - y^{2})

\int \frac{x ^{3}}{x ^{2} - 36} d x

\frac{\partial}{\partial x} (e^{2 x} \cdot cos (x + y))

y = ∣ 2 x ∣, y = x^{2} - 3

What is the integral of (-x^2+6x-5) ?
The integral of (-x^2+6x-5) is -(x^3)/3+3x^2-5x+C