We use the Fundamental Theorem of Integral Calculus
#: intf(t)dt=F(t) rArr ""int_a^bf(t)dt=F(b)-F(a).#
So, we first find out the Indefinite Integral #I=intsqrttlntdt,# using the
Method of Integration by Parts,which states that,
#intuvdt=uintvdt-int{(du)/dtintvdt}dt.#
We apply this by taking #u=lnt, &, v=sqrtt,# so that,
#(du)/dt=1/t, &, intvdt=(t^(3/2))/(3/2)=(2/3)t^(3/2).#
#:. I=(2/3)t^(3/2)lnt-int{1/t*(2/3)t^(3/2)}dt,#
#=(2/3)t^(3/2)lnt-(2/3)intt^(1/2)dt,#
#=(2/3)t^(3/2)lnt-(2/3)(2/3)t^(3/2)#,
#:. I=(2/3)t^(3/2)lnt-(4/9)t^(3/2)+C.#
Hence, #int_2^1intsqrttlntdt=[(2/3)*1^(3/2)*ln1-(4/9)*1^(3/2)]-[(2/3)*2^(3/2)*ln2-(4/9)*2^(3/2)]#
#=[0-4/9]-[(2^(5/2)*ln2)/3-4/9*2^(3/2)]#
#=4/9(2^(3/2)-1)-2^(5/2)/3*(ln2)#
#=4/9(2sqrt2-1)-(4/3sqrt2)ln2.#