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

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

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

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

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

Solution steps

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

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

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

\int 2 d x = 2 x

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

Add a constant to the solution

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

\frac{d}{d θ} (csc (θ))

x \to \infty lim (- 2^{x})

x ln (x) = y (1 + 15 + y^{2}) y^{^{'}} (x), y (1) = 1

\int e^{12 x} \cdot x^{3} d x

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