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

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

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

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

3 x - x^{2} - \frac{x ^{5}}{5} + C

Solution steps

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

\int 3 d x = 3 x

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

\int x^{4} d x = \frac{x ^{5}}{5}

= 3 x - x^{2} - \frac{x ^{5}}{5}

Add a constant to the solution

= 3 x - x^{2} - \frac{x ^{5}}{5} + C

\int 10 sin (x) 2 x d x

\frac{d}{d x} (arcsec (x) + arccsc (x))

\frac{x}{x + 3}

y^{^{''}} - y = 0, y (0) = 1, y^{^{'}} (0) = 2

y = \frac{x ^{2} - 7 x + 2}{x - 2}

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