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

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

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

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

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

Solution steps

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

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

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{9}{( n + 1 ) !}

f (x) = 8 x^{2}, at x = 10

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

\int 1 + tan^{2} (x) - 2 cos (x) d x

\frac{\partial}{\partial r} (\frac{π}{3} h^{2} (3 r - h))

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