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

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

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

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

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

Solution steps

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

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

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

\int 4 d x = 4 x

= \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} - 4 x

Add a constant to the solution

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

f (x) = x + 4, (12, 4)

4 y^{^{''}} + 6 y^{^{'}} - 4 y = 0

\int 1 - cos (x) d x

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

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

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