What is the integral of (x^3-6x+18)/(3x^{1/2)} ?

The integral of (x^3-6x+18)/(3x^{1/2)} is 1/3 (2/7 x^{7/2}-4x^{3/2}+36x^{1/2})+C

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integral of (x^3-6x+18)/(3x^{1/2)}

\int \frac{x ^{3} - 6 x + 18}{3 x ^{\frac{1}{2}}} d x

\frac{1}{3} (\frac{2}{7} x^{\frac{7}{2}} - 4 x^{\frac{3}{2}} + 36 x^{\frac{1}{2}}) + C

Solution steps

\int \frac{x ^{3} - 6 x + 18}{3 x ^{\frac{1}{2}}} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{x ^{3} - 6 x + 18}{x ^{\frac{1}{2}}} d x

Expand

\frac{x ^{3} - 6 x + 18}{x ^{\frac{1}{2}}} : x^{\frac{5}{2}} - 6 x^{\frac{1}{2}} + \frac{18}{x ^{\frac{1}{2}}}

= \frac{1}{3} \cdot \int x^{\frac{5}{2}} - 6 x^{\frac{1}{2}} + \frac{18}{x ^{\frac{1}{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}{3} (\int x^{\frac{5}{2}} d x - \int 6 x^{\frac{1}{2}} d x + \int \frac{18}{x ^{\frac{1}{2}}} d x)

\int x^{\frac{5}{2}} d x = \frac{2}{7} x^{\frac{7}{2}}

\int 6 x^{\frac{1}{2}} d x = 4 x^{\frac{3}{2}}

\int \frac{18}{x ^{\frac{1}{2}}} d x = 36 x^{\frac{1}{2}}

= \frac{1}{3} (\frac{2}{7} x^{\frac{7}{2}} - 4 x^{\frac{3}{2}} + 36 x^{\frac{1}{2}})

Add a constant to the solution

= \frac{1}{3} (\frac{2}{7} x^{\frac{7}{2}} - 4 x^{\frac{3}{2}} + 36 x^{\frac{1}{2}}) + C

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What is the integral of (x^3-6x+18)/(3x^{1/2)} ?
The integral of (x^3-6x+18)/(3x^{1/2)} is 1/3 (2/7 x^{7/2}-4x^{3/2}+36x^{1/2})+C