What is the integral of (6+x+x^2)/(sqrt(x)) ?

The integral of (6+x+x^2)/(sqrt(x)) is 12sqrt(x)+2/3 x^{3/2}+2/5 x^{5/2}+C

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integral of (6+x+x^2)/(sqrt(x))

\int \frac{6 + x + x ^{2}}{x} d x

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

Solution steps

\int \frac{6 + x + x ^{2}}{x} d x

Expand

\frac{6 + x + x ^{2}}{x} : \frac{6}{x} + x + x^{\frac{3}{2}}

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

\int \frac{6}{x} d x = 12 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}}

= 12 x + \frac{2}{3} x^{\frac{3}{2}} + \frac{2}{5} x^{\frac{5}{2}}

Add a constant to the solution

= 12 x + \frac{2}{3} x^{\frac{3}{2}} + \frac{2}{5} x^{\frac{5}{2}} + C

\frac{d x}{d t} = - 3 x + 6 e^{- t}

\frac{d}{d x} (x^{3} - 4 x^{2} + 45 x + 10)

\int_{0}^{5} x e^{- 2 x} d x

y^{^{'}} = e^{4 x} - 6 x

x \to 0 lim (\frac{- 6}{e ^{3} - x})

What is the integral of (6+x+x^2)/(sqrt(x)) ?
The integral of (6+x+x^2)/(sqrt(x)) is 12sqrt(x)+2/3 x^{3/2}+2/5 x^{5/2}+C