We begin with the substitution #x= tan theta#. Then #dx = sec^2 theta d theta# and #sqrt(x^2+1) = sec theta# Thus
#int_(3^(1/2))^(35^(1/2))(x^2+1)^(1/2)/xdx = int_(theta_1)^(theta_2) sec theta/tan theta sec^2 theta d theta #
where #theta_1 = tan^-1sqrt3# and #theta_2 = tan^-1sqrt35#
This leads to
#int_(theta_1)^(theta_2) sec^3 theta/tan theta d theta = int_(theta_1)^(theta_2) (d theta)/(sin theta cos^2 theta)#
#qquadqquad = int_(theta_1)^(theta_2) (sin theta d theta)/(sin^2 theta cos^2 theta)#
We next substitute #cos theta =u # and use the following
- #sin theta d theta = -du#
- # sin^2 theta = 1-u^2#
- #sec theta_1 = sqrt(tan^2theta_1+1) = 2 implies cos theta_1 = 1/2#
- #sec theta_2 = sqrt(tan^2theta_2+1) = 6 implies cos theta_2 = 1/6#
to reduce our integral to
#int_(1/2)^(1/6) (-du)/((1-u^2)u^2) = int_(1/6)^(1/2)(1/(1-u^2)+1/u^2)du#
#qquad=[1/2 ln((1+u)/(1-u))-1/u]_(1/6)^(1/2)#
#qquad=1/2 {ln((1+1/2)/(1-1/2))-ln((1+1/6)/(1-1/6))}-1/(1/2)+1/(1/6)#
#qquad=1/2{ln(3)-ln(7/5)}-2+6#
#qquad = 4+1/2ln(15/7)#
