What is the integral of (3x^2-4x+7) ?

The integral of (3x^2-4x+7) is x^3-2x^2+7x+C

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integral of (3x^2-4x+7)

\int (3 x^{2} - 4 x + 7) d x

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

Solution steps

\int 3 x^{2} - 4 x + 7 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 3 x^{2} d x - \int 4 x d x + \int 7 d x

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

\int 4 x d x = 2 x^{2}

\int 7 d x = 7 x

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

Add a constant to the solution

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

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

\frac{d y}{d x} = \frac{( x y + 3 x - y - 3 )}{( x y - 2 x + 4 y - 8 )}

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

\frac{d}{d x} (\frac{2 - 5 x}{3 x ^{\frac{1}{3}}})

x \to \infty lim (\frac{( 25 x + 7 ) 5 x ^{2} + 2}{5 x ^{2} + 2})

What is the integral of (3x^2-4x+7) ?
The integral of (3x^2-4x+7) is x^3-2x^2+7x+C