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

The integral of (x^3-2x^4+x^5)/2 is (x^4)/8-(x^5)/5+(x^6)/(12)+C

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

\int \frac{x ^{3} - 2 x ^{4} + x ^{5}}{2} d x

\frac{x ^{4}}{8} - \frac{x ^{5}}{5} + \frac{x ^{6}}{12} + C

Solution steps

\int \frac{x ^{3} - 2 x ^{4} + x ^{5}}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int x^{3} - 2 x^{4} + x^{5} 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}{2} (\int x^{3} d x - \int 2 x^{4} d x + \int x^{5} d x)

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

\int 2 x^{4} d x = \frac{2 x ^{5}}{5}

\int x^{5} d x = \frac{x ^{6}}{6}

= \frac{1}{2} (\frac{x ^{4}}{4} - \frac{2 x ^{5}}{5} + \frac{x ^{6}}{6})

Simplify

\frac{1}{2} (\frac{x ^{4}}{4} - \frac{2 x ^{5}}{5} + \frac{x ^{6}}{6}) : \frac{x ^{4}}{8} - \frac{x ^{5}}{5} + \frac{x ^{6}}{12}

= \frac{x ^{4}}{8} - \frac{x ^{5}}{5} + \frac{x ^{6}}{12}

Add a constant to the solution

= \frac{x ^{4}}{8} - \frac{x ^{5}}{5} + \frac{x ^{6}}{12} + C

3 tan (x)

f (x) = 12 x, (9, 36)

n = 0 \sum \infty \frac{1}{2} \cdot (- 1)^{n}

\frac{\partial}{\partial x} (- 4 x y - x^{4} - y^{4})

f (x) = x + 4

What is the integral of (x^3-2x^4+x^5)/2 ?
The integral of (x^3-2x^4+x^5)/2 is (x^4)/8-(x^5)/5+(x^6)/(12)+C