Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 26

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 .