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

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

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

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

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

Solution steps

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

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

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

\int 5 d x = 5 x

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

Add a constant to the solution

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

f (x) = - \frac{3}{x}

\int \frac{sin ( 2 x )}{32 + cos ^{2} ( x )} d x

\frac{d P}{d t} = 0.16 P (1 - \frac{P}{1000}) - 30, P (0) = 200

6 cos (t) - 2 sin (t)

f (t) = t^{3} - 27 t + 5

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