By parts.
INT.[u]dv = u*v -INT.[v]du -------(i)
INT.[sqrt(y) *ln(y)]dy -------------(1)
You can try two ways. I did, and I found this is the correct assumption.
Let u = ln(y)
Hence, du = (1/y)dy
And so, dv = sqrt(y) dy = (y^(1/2))dy
Hence, v = 1/(3/2) *y^(3/2) = (2/3)y^(3/2)
INT.[sqrt(y) *ln(y)]dy -------------(1)
Apply the assumptions, and (i),
= ln(y)*[(2/3)y^(3/2)] -INT.[(2/3)y^(3/2)][(1/y)dy]
= (2/3)(y^(3/2))ln(y) -(2/3)INT.[y^(1/2)]dy
= (2/3){y^(3/2) ln(y) -(2/3)(y^(3/2))} +C
= [(2/3)y^(3/2)]{ln(y) -2/3} +C
= (2/3)(y)(sqrt(y))[ln(y) -2/3] +C --------------answer.
= [ln(y) *(2/3)y^(3/2)] -