# How do you find the integral of int (lnx / sqrt(x)) dx from 0 to 1?

Sep 15, 2015

I found: $- 4$

#### Explanation:

Let us try by changing $\frac{1}{\sqrt{x}} = = {x}^{- \frac{1}{2}}$ and then Integrate By Parts:
$\int \left(\ln \frac{x}{\sqrt{x}}\right) \mathrm{dx} = = \int \left({x}^{- \frac{1}{2}} \ln \left(x\right)\right) \mathrm{dx} =$
$= 2 {x}^{\frac{1}{2}} \ln \left(x\right) - \int \left(2 {x}^{\frac{1}{2}} \cdot \frac{1}{x} \mathrm{dx}\right) =$
$= 2 {x}^{\frac{1}{2}} \ln \left(x\right) - \int \left(2 {x}^{\frac{1}{2} - 1} \mathrm{dx}\right) =$
$= 2 {x}^{\frac{1}{2}} \ln \left(x\right) - \int \left(2 {x}^{- \frac{1}{2}} \mathrm{dx}\right) =$
$= 2 {x}^{\frac{1}{2}} \ln \left(x\right) - 4 {x}^{\frac{1}{2}} =$
let us use our extrema:
$= 2 {x}^{\frac{1}{2}} \ln \left(x\right) - 4 {x}^{\frac{1}{2}} {|}_{0}^{1}$$= - 4$

Only problematic part is $2 {x}^{\frac{1}{2}} \ln x$ when $x \to {0}_{+}$, so we will find the limit:

${\lim}_{x \to {0}_{+}} 2 {x}^{\frac{1}{2}} \ln x = 0 \cdot \left(- \infty\right)$ which is undefined.

Using the rule of L'Hospital:

${\lim}_{x \to {0}_{+}} 2 {x}^{\frac{1}{2}} \ln x = 2 {\lim}_{x \to {0}_{+}} \ln \frac{x}{x} ^ \left(- \frac{1}{2}\right) =$

$= 2 {\lim}_{x \to {0}_{+}} \frac{\frac{1}{x}}{- \frac{1}{2} {x}^{- \frac{3}{2}}} = - 4 {\lim}_{x \to {0}_{+}} {x}^{\frac{3}{2}} / x =$

$= - 4 {\lim}_{x \to {0}_{+}} \sqrt{x} = - 4 \cdot 0 = 0$

So, the above results holds.

Sep 16, 2015

i got -4 by changing variabe method

#### Explanation:

let the $x = {t}^{2}$
then ${x}^{\frac{1}{2}} = t$
$\mathrm{dx}$ = $2 t \mathrm{dt}$
limits:
if x= 0 then t= 0
if x= 1 then t=1
applying integration
${\int}_{0}^{1}$ $\ln {t}^{2} / t$( $2 t \mathrm{dt}$)
=${\int}_{0}^{1}$$4 \ln t \mathrm{dt}$
=4${\int}_{0}^{1}$$\ln t \mathrm{dt}$
applying by parts
4$\ln t$*${\int}_{0}^{1}$$\mathrm{dt}$-${\int}_{0}^{1}$$t \cdot \left(\frac{1}{t}\right)$$\mathrm{dt}$
=4[$\ln 1$ 1- 0 $\ln 0$-{1-0}]
=4[0-1]
=-4