What is the integral of (3-2x-4x^2)(1+4x) ?

The integral of (3-2x-4x^2)(1+4x) is 3x+5x^2-4x^3-4x^4+C

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integral of (3-2x-4x^2)(1+4x)

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

3 x + 5 x^{2} - 4 x^{3} - 4 x^{4} + C

Solution steps

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

Expand

(3 - 2 x - 4 x^{2}) (1 + 4 x) : 3 + 10 x - 12 x^{2} - 16 x^{3}

= \int 3 + 10 x - 12 x^{2} - 16 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 d x + \int 10 x d x - \int 12 x^{2} d x - \int 16 x^{3} d x

\int 3 d x = 3 x

\int 10 x d x = 5 x^{2}

\int 12 x^{2} d x = 4 x^{3}

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

= 3 x + 5 x^{2} - 4 x^{3} - 4 x^{4}

Add a constant to the solution

= 3 x + 5 x^{2} - 4 x^{3} - 4 x^{4} + C

\int 10 x sin (x) d x

y = 5 + 4 x^{2}

\frac{\partial}{\partial x} (12 x^{2} y^{3})

x \to 0 + lim ((e^{x} + x)^{\frac{3}{x}})

\int \frac{1}{x ^{2} + 2 x + 3} d x

What is the integral of (3-2x-4x^2)(1+4x) ?
The integral of (3-2x-4x^2)(1+4x) is 3x+5x^2-4x^3-4x^4+C