# How do you integrate int lnx/sqrtx by integration by parts method?

Jul 23, 2016

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

#### Explanation:

We derive the IBP formula from the product rule:

Take functions $u \left(x\right) , v \left(x\right)$

$\left(u v\right) ' = u ' v + u v '$

$\implies u v ' = \left(u v\right) ' - u ' v$

$\therefore \int u v ' \mathrm{dx} = u v - \int u ' v \mathrm{dx}$

In this case, we have to take $u \left(x\right) = \ln \left(x\right) \mathmr{and} v \left(x\right) = {x}^{- \frac{1}{2}}$ because we cannot integrate the natural log function directly.

Obtain:

$u ' \left(x\right) = \frac{1}{x} \mathmr{and} v \left(x\right) = 2 {x}^{\frac{1}{2}}$

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

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

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

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