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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Add a constant to the solution

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

f (x) = ln (9 x)

f (x) = - 4 x^{2} + 6, at x = - 1

\frac{d y}{d x} = - \frac{( x e ^{- x^{2}} )}{y}

\int x^{(\frac{3}{2})} d x

\frac{d}{d x} (arcsin (x) \frac{8}{π})

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