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

The integral of ((x^2-x^3))/(2sqrt(x)) is 1/35 x^{5/2}(-5x+7)+C

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integral of ((x^2-x^3))/(2sqrt(x))

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

\frac{1}{35} x^{\frac{5}{2}} (- 5 x + 7) + C

Solution steps

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

Expand

\frac{x ^{2} - x ^{3}}{x} : \frac{x ^{2}}{x} - \frac{x ^{3}}{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 \frac{x ^{2}}{x} d x - \int \frac{x ^{3}}{x} d x)

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

\int \frac{x ^{3}}{x} d x = \frac{2}{7} x^{\frac{7}{2}}

= \frac{1}{2} (\frac{2}{5} x^{\frac{5}{2}} - \frac{2}{7} x^{\frac{7}{2}})

Simplify

\frac{1}{2} (\frac{2}{5} x^{\frac{5}{2}} - \frac{2}{7} x^{\frac{7}{2}}) : \frac{1}{35} x^{\frac{5}{2}} (- 5 x + 7)

= \frac{1}{35} x^{\frac{5}{2}} (- 5 x + 7)

Add a constant to the solution

= \frac{1}{35} x^{\frac{5}{2}} (- 5 x + 7) + C

\int - \frac{4}{t} d t

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\int \frac{1}{( w ^{2} + 5 ) ^{3}} d w

\int \frac{1}{y y ^{2} - 16} d y

\int \frac{2 + x}{1 + x ^{2}} d x

What is the integral of ((x^2-x^3))/(2sqrt(x)) ?
The integral of ((x^2-x^3))/(2sqrt(x)) is 1/35 x^{5/2}(-5x+7)+C