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

x \to 4 - lim (\frac{6 x}{x - 4})

f (x) = 3 sin^{2} (x)

2 y^{^{''}} + y^{^{'}} - 2 y = 0

x \to 5 lim (cos (x))

y = 2 x^{3 π}

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