# How do you integrate int (x^2+x+1)/sqrtxdx?

Dec 20, 2016

$\int \frac{{x}^{2} + x + 1}{\sqrt{x}} \mathrm{dx} = \frac{\frac{2}{5} {x}^{3} + \frac{2}{3} {x}^{2} + 2 x}{\sqrt{x}}$

#### Explanation:

As the integral is a linear operator, we can integrate term by term:

$\int \frac{{x}^{2} + x + 1}{\sqrt{x}} \mathrm{dx} = \int {x}^{2} / \sqrt{x} \mathrm{dx} + \int \frac{x}{\sqrt{x}} \mathrm{dx} + \int \frac{1}{\sqrt{x}} \mathrm{dx} = \int {x}^{\frac{3}{2}} \mathrm{dx} + \int {x}^{\frac{1}{2}} \mathrm{dx} + \int {x}^{- \frac{1}{2}} \mathrm{dx} = \frac{2}{5} {x}^{\frac{5}{2}} + \frac{2}{3} {x}^{\frac{3}{2}} + 2 {x}^{\frac{1}{2}} = \frac{\frac{2}{5} {x}^{3} + \frac{2}{3} {x}^{2} + 2 x}{\sqrt{x}}$

Dec 20, 2016

The answer is $= 2 \left({x}^{\frac{5}{2}} / 5 + {x}^{\frac{3}{2}} / 3 + \sqrt{x}\right) + C$

#### Explanation:

We need

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(n \ne - 1\right)$

Therefore,

$\int \frac{\left({x}^{2} + x + 1\right) \mathrm{dx}}{\sqrt{x}}$

$= \int \left({x}^{2} / \sqrt{x} + \frac{x}{\sqrt{x}} + \frac{1}{\sqrt{x}}\right) \mathrm{dx}$

$= \int \left({x}^{\frac{3}{2}} + \sqrt{x} + {x}^{- \frac{1}{2}}\right) \mathrm{dx}$

$= {x}^{\frac{5}{2}} / \left(\frac{5}{2}\right) + {x}^{\frac{3}{2}} / \left(\frac{3}{2}\right) + {x}^{\frac{1}{2}} / \left(\frac{1}{2}\right) + C$

$= 2 \left({x}^{\frac{5}{2}} / 5 + {x}^{\frac{3}{2}} / 3 + {x}^{\frac{1}{2}}\right) + C$