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

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

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

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

- \frac{1}{4} (\frac{x ^{5}}{5} + 2 x) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{4} \cdot \int x^{4} + 2 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

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

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

\int 2 d x = 2 x

= - \frac{1}{4} (\frac{x ^{5}}{5} + 2 x)

Add a constant to the solution

= - \frac{1}{4} (\frac{x ^{5}}{5} + 2 x) + C

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

\int 18 x \cdot e^{3 x} d x

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\int 4 + 2 r^{2} r d r

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

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