What is the integral of (3x^2-4x+5) ?

The integral of (3x^2-4x+5) is x^3-2x^2+5x+C

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integral of (3x^2-4x+5)

\int (3 x^{2} - 4 x + 5) d x

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

Solution steps

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

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

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

\int 5 d x = 5 x

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

Add a constant to the solution

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

(5 cos (t))^{^{'}}

\frac{d}{d x} (sin (5 x^{2}))

x \to 1 lim (\frac{\frac{1}{x} - 1}{x - 1})

y = \frac{6}{x ^{4}}

x \to 0 lim (x^{3} (ln (x)))

What is the integral of (3x^2-4x+5) ?
The integral of (3x^2-4x+5) is x^3-2x^2+5x+C