Question

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

Answer 1

I found: `-4`

Explanation:

Let us try by changing `1/sqrt(x)==x^(-1/2)` and then Integrate By Parts:
`int(ln(x)/sqrt(x))dx==int(x^(-1/2)ln(x))dx=`
`=2x^(1/2)ln(x)-int(2x^(1/2)*1/xdx)=`
`=2x^(1/2)ln(x)-int(2x^(1/2-1)dx)=`
`=2x^(1/2)ln(x)-int(2x^(-1/2)dx)=`
`=2x^(1/2)ln(x)-4x^(1/2)=`
let us use our extrema:
`=2x^(1/2)ln(x)-4x^(1/2)|_0^1` `=-4`

Only problematic part is `2x^(1/2)lnx` when `x->0_+`, so we will find the limit:

`lim_(x->0_+)2x^(1/2)lnx= 0*(-oo)` which is undefined.

Using the rule of L'Hospital:

`lim_(x->0_+)2x^(1/2)lnx=2lim_(x->0_+)lnx/x^(-1/2)=`

`=2lim_(x->0_+)(1/x)/(-1/2x^(-3/2))=-4lim_(x->0_+)x^(3/2)/x=`

`=-4lim_(x->0_+)sqrtx=-4*0=0`

So, the above results holds.

Answer 2

i got -4 by changing variabe method

Explanation:

let the `x = t^2`
then `x^(1/2) = t`
`dx` = `2tdt`
limits:
if x= 0 then t= 0
if x= 1 then t=1
applying integration
`int_0^1` `lnt^2/t`( `2tdt`)
=`int_0^1` `4lntdt`
=4`int_0^1` `lntdt`
applying by parts
4`lnt`*`int_0^1` `dt`-`int_0^1` `t*(1/t)` `dt`
=4[`ln1` 1- 0 `ln0`-{1-0}]
=4[0-1]
=-4