What is the integral of (21x^2-60x+17) ?

The integral of (21x^2-60x+17) is 7x^3-30x^2+17x+C

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integral of (21x^2-60x+17)

\int (21 x^{2} - 60 x + 17) d x

7 x^{3} - 30 x^{2} + 17 x + C

Solution steps

\int 21 x^{2} - 60 x + 17 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 21 x^{2} d x - \int 60 x d x + \int 17 d x

\int 21 x^{2} d x = 7 x^{3}

\int 60 x d x = 30 x^{2}

\int 17 d x = 17 x

= 7 x^{3} - 30 x^{2} + 17 x

Add a constant to the solution

= 7 x^{3} - 30 x^{2} + 17 x + C

\frac{\partial}{\partial x} (x^{2} y + 3 e^{x y})

\int 2 (u + 5) + 1 (u) d u

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

\frac{d}{d x} (x^{2} y^{3})

(t - 2)^{2}

What is the integral of (21x^2-60x+17) ?
The integral of (21x^2-60x+17) is 7x^3-30x^2+17x+C