What is the integral of ((1+x+x^2))/(sqrt(x)) ?

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

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integral of ((1+x+x^2))/(sqrt(x))

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

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

Solution steps

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

Expand

\frac{1 + x + x ^{2}}{x} : \frac{1}{x} + x + x^{\frac{3}{2}}

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

\int \frac{1}{x} d x = 2 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}}

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

Add a constant to the solution

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

x \to 2 lim (\frac{2}{x ^{2} - 4})

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

y^{^{''}} - 9 y^{^{'}} = 1 + 3 e^{2 x}

\frac{d y}{d x} = \frac{y}{x + 1}

t \to 0 lim (sin (5 t))

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