# How do you find the integral ln(x)/sqrtx?

Aug 28, 2015

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

#### Explanation:

Use integration-by-parts. Let $u = \ln \left(x\right)$ so that $\mathrm{du} = \frac{1}{x} \setminus \mathrm{dx}$ and $\mathrm{dv} = \frac{1}{\sqrt{x}} \setminus \mathrm{dx} = {x}^{- \frac{1}{2}} \setminus \mathrm{dx}$ so that $v = 2 {x}^{\frac{1}{2}} = 2 \sqrt{x}$.

The integration-by-parts formula can be written in abstract form as: $\int u \setminus \mathrm{dv} = u v - \int v \setminus \mathrm{du}$.

For this problem, this becomes:

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

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

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

$= 2 \sqrt{x} \ln \left(x\right) - 4 \sqrt{x} + C$