# How do you find the integral ln x / x^(1/2)?

Apr 10, 2018

$\int \ln \frac{x}{x} ^ \left(\frac{1}{2}\right) \mathrm{dx} = 2 \sqrt{x} \ln x - 4 \sqrt{x} + C$

#### Explanation:

Rewrite the integrand, and integrate by parts:

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

$u = \ln x$

$\mathrm{du} = {x}^{-} 1 \mathrm{dx}$

$\mathrm{dv} = {x}^{- \frac{1}{2}} \mathrm{dx}$

$v = \int {x}^{- \frac{1}{2}} \mathrm{dx} = 2 {x}^{\frac{1}{2}}$

$u v - \int v \mathrm{du} = 2 \sqrt{x} \ln x - 2 \int {x}^{\frac{1}{2}} {x}^{-} 1 \mathrm{dx}$

$= 2 \sqrt{x} \ln x - 2 \int {x}^{- \frac{1}{2}} \mathrm{dx} = 2 \sqrt{x} \ln x - 4 \sqrt{x} + C$

So,

$\int \ln \frac{x}{x} ^ \left(\frac{1}{2}\right) \mathrm{dx} = 2 \sqrt{x} \ln x - 4 \sqrt{x} + C$

Apr 10, 2018

$2 \sqrt{x} I n x - 4 \sqrt{x} + C$

#### Explanation:

Let $u = I n x$
and $\mathrm{dv} = {x}^{- \frac{1}{2}}$ SO $v = 2 \sqrt{x}$

Integration by parts = $u v - \int v \mathrm{du}$

$\int \frac{I n x}{x} ^ \left(\frac{1}{2}\right) \mathrm{dx}$ = $2 \sqrt{x} \times I n x - \int 2 \sqrt{x} \times \frac{1}{x}$

=$2 \sqrt{x} I n x - 2 \int {x}^{- \frac{1}{2}} \mathrm{dx}$

=$2 \sqrt{x} I n x - 2 \times 2 \sqrt{x} + C$

=$2 \sqrt{x} I n x - 4 \sqrt{x} + C$