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

Expert Answers
lemjay eNotes educator| Certified Educator

`int_4^9 lny/sqrty dy`

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

So let:

`u = ln y `    


`dv=int1/sqrty dy`

Then, differentiate u and integrate dv.

`du=1/y dy`


`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`


`=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)`




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


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




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