What is the integral of (16x^3-3x^2+2) ?

The integral of (16x^3-3x^2+2) is 4x^4-x^3+2x+C

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integral of (16x^3-3x^2+2)

\int (16 x^{3} - 3 x^{2} + 2) d x

4 x^{4} - x^{3} + 2 x + C

Solution steps

\int 16 x^{3} - 3 x^{2} + 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 16 x^{3} d x - \int 3 x^{2} d x + \int 2 d x

\int 16 x^{3} d x = 4 x^{4}

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

\int 2 d x = 2 x

= 4 x^{4} - x^{3} + 2 x

Add a constant to the solution

= 4 x^{4} - x^{3} + 2 x + C

x \to 2 lim (\frac{3 x ^{2} + 4 - 4 x ^{2} + 6 x}{x ^{2} - 4})

\int \frac{cos ( x )}{3 + sin ( x )} d x

\int \frac{x ^{3} + 4 x - 2}{( x - 2 ) ( x + 1 )} d x

f (x) = \frac{x ^{3}}{4}, at x = - 6

\int tan^{4} (x) sec^{6} (x) d x

What is the integral of (16x^3-3x^2+2) ?
The integral of (16x^3-3x^2+2) is 4x^4-x^3+2x+C