What is the integral of 5x(sqrt(x)-x^2) ?

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

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integral of 5x(sqrt(x)-x^2)

\int 5 x (x - x^{2}) d x

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

Solution steps

\int 5 x (x - x^{2}) d x

Take the constant out:

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

= 5 \cdot \int x (x - x^{2}) d x

Expand

x (x - x^{2}) : x^{\frac{3}{2}} - x^{3}

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

= 5 (\int x^{\frac{3}{2}} d x - \int x^{3} d x)

\int x^{\frac{3}{2}} d x = \frac{2}{5} x^{\frac{5}{2}}

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

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

Simplify

5 (\frac{2}{5} x^{\frac{5}{2}} - \frac{x ^{4}}{4}) : 2 x^{\frac{5}{2}} - \frac{5 x ^{4}}{4}

= 2 x^{\frac{5}{2}} - \frac{5 x ^{4}}{4}

Add a constant to the solution

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

\frac{\partial}{\partial x} (ln (x^{2} + y) - z)

x \to 3 lim (x^{2} + 9)

2 sin^{2} (x)

f (y) = \frac{1}{2 y}

\frac{d}{d x} (9 arctan (x - 1 + x^{2}))

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