What is the integral of 4x^3-20x^2+8x+32 ?

The integral of 4x^3-20x^2+8x+32 is x^4-(20x^3)/3+4x^2+32x+C

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integral of 4x^3-20x^2+8x+32

\int 4 x^{3} - 20 x^{2} + 8 x + 32 d x

x^{4} - \frac{20 x ^{3}}{3} + 4 x^{2} + 32 x + C

Solution steps

\int 4 x^{3} - 20 x^{2} + 8 x + 32 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 4 x^{3} d x - \int 20 x^{2} d x + \int 8 x d x + \int 32 d x

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

\int 20 x^{2} d x = \frac{20 x ^{3}}{3}

\int 8 x d x = 4 x^{2}

\int 32 d x = 32 x

= x^{4} - \frac{20 x ^{3}}{3} + 4 x^{2} + 32 x

Add a constant to the solution

= x^{4} - \frac{20 x ^{3}}{3} + 4 x^{2} + 32 x + C

\frac{d}{d x} (- 2 sin (3 x))

f (x) = \frac{5}{x}, at x = 9

\frac{d}{d x} (\frac{cos ( x )}{x ^{8}})

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

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

What is the integral of 4x^3-20x^2+8x+32 ?
The integral of 4x^3-20x^2+8x+32 is x^4-(20x^3)/3+4x^2+32x+C