What is the integral of (8+x+x^2)/(sqrt(x)) ?

The integral of (8+x+x^2)/(sqrt(x)) is 16sqrt(x)+2/3 x^{3/2}+2/5 x^{5/2}+C

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integral of (8+x+x^2)/(sqrt(x))

\int \frac{8 + x + x ^{2}}{x} d x

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

Solution steps

\int \frac{8 + x + x ^{2}}{x} d x

Expand

\frac{8 + x + x ^{2}}{x} : \frac{8}{x} + x + x^{\frac{3}{2}}

= \int \frac{8}{x} + x + x^{\frac{3}{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

= \int \frac{8}{x} d x + \int x d x + \int x^{\frac{3}{2}} d x

\int \frac{8}{x} d x = 16 x

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

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

= 16 x + \frac{2}{3} x^{\frac{3}{2}} + \frac{2}{5} x^{\frac{5}{2}}

Add a constant to the solution

= 16 x + \frac{2}{3} x^{\frac{3}{2}} + \frac{2}{5} x^{\frac{5}{2}} + C

\int_{0}^{\frac{1}{2}} 5 arccos (x) d x

\frac{d}{d x} (e^{a x^{6}})

f (x) = (x)^{3} \cdot e^{(3 x)}

2 y^{2} = \frac{d y}{d x}

What is the integral of (8+x+x^2)/(sqrt(x)) ?
The integral of (8+x+x^2)/(sqrt(x)) is 16sqrt(x)+2/3 x^{3/2}+2/5 x^{5/2}+C