What is the integral of (9x^2-5x+4) ?

The integral of (9x^2-5x+4) is 3x^3-(5x^2)/2+4x+C

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integral of (9x^2-5x+4)

\int (9 x^{2} - 5 x + 4) d x

3 x^{3} - \frac{5 x ^{2}}{2} + 4 x + C

Solution steps

\int 9 x^{2} - 5 x + 4 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 9 x^{2} d x - \int 5 x d x + \int 4 d x

\int 9 x^{2} d x = 3 x^{3}

\int 5 x d x = \frac{5 x ^{2}}{2}

\int 4 d x = 4 x

= 3 x^{3} - \frac{5 x ^{2}}{2} + 4 x

Add a constant to the solution

= 3 x^{3} - \frac{5 x ^{2}}{2} + 4 x + C

\frac{d}{d x} (sin^{2} (2 x) - cos (2 x))

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

x \to 5 lim (4 x + ∣ x - 5 ∣)

n = 1 \sum \infty \frac{x ^{n}}{1 2 ^{n}}

\frac{d}{d t} (cos (t))

What is the integral of (9x^2-5x+4) ?
The integral of (9x^2-5x+4) is 3x^3-(5x^2)/2+4x+C