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

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

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

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

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

Solution steps

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

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

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

\int 4 d x = 4 x

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

Add a constant to the solution

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

\frac{d}{d x} (csc (6 x))

2 x^{4} - 10 x^{2}

\int_{\frac{π}{2}}^{π} π sin^{2} (x) d x

f (x) = 4 x^{2} + 2 x + 5, at x = 3

t (\frac{d y}{d t}) + 33 y = \frac{1}{t}

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