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

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

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integral of sqrt(x)-x^4

\int x - x^{4} d x

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

Solution steps

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{5 x - 2}{2 x + 7})

y^{^{''}} - 3 y^{^{'}} - 28 y = 0

x \to 0 lim (\frac{e ^{x^{6}} - 1}{x})

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

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

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