`int_4^9 ln(y)/sqrt(y) dy` Evaluate the integral

`int_4^9 lny/sqrty dy`

To evaluate, apply integration by parts `intudv=uv-intvdu` .

So let:

`u = ln y `    

and    

`dv=int1/sqrty dy`

Then, differentiate u and integrate dv.

`du=1/y dy`

and

`v=int 1/sqrty dy = int y^(-1/2)=2y^(1/2)`

Plug-in them to the formula. So the integral becomes:

`int lny/(sqrty)dy`

`= lny*2y^(1/2) - 2y^(1/2)*1/ydy`

`=2y^(1/2)lny - 2int y^(-1/2) dy`

`=2y^(1/2)lny-2*2y^(1/2)`

`=2sqrtylny - 4sqrty`

And, substitute the limits of the integral.

`int _4^9 lny/ydy`

`=...

`int_4^9 lny/sqrty dy`

To evaluate, apply integration by parts `intudv=uv-intvdu` .

So let:

`u = ln y `    

and    

`dv=int1/sqrty dy`

Then, differentiate u and integrate dv.

`du=1/y dy`

and

`v=int 1/sqrty dy = int y^(-1/2)=2y^(1/2)`

Plug-in them to the formula. So the integral becomes:

`int lny/(sqrty)dy`

`= lny*2y^(1/2) - 2y^(1/2)*1/ydy`

`=2y^(1/2)lny - 2int y^(-1/2) dy`

`=2y^(1/2)lny-2*2y^(1/2)`

`=2sqrtylny - 4sqrty`

And, substitute the limits of the integral.

`int _4^9 lny/ydy`

`= (2sqrtylny - 4sqrty)|_4^9`

`=(2sqrt9 ln9 - 4sqrt9)-(2sqrt4ln4-4sqrt4)`

`=(2*3ln9 - 4*3)-(2*2ln4-4*2)`

`=(6ln9-12)-(4ln4-8)`

`=6ln9-4ln4-4`

`=ln(9^6)-ln(4^4)-4`

`=ln (9^6/4^4)-4`

`=ln(((3^2)^6)/(4^4))-4`

`=ln (((3^3)^4)/(4^4))-4`

`=ln((3^3/4)^4)-4`

`=4ln(27/4)-4`

Therefore, `int_4^9 lny/sqrtydy = 4ln (27/4)-4` .