We begin with the substitution x= tan thetax=tanθ. Then dx = sec^2 theta d thetadx=sec2θdθ and sqrt(x^2+1) = sec theta√x2+1=secθ 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 ∫3512312(x2+1)12xdx=∫θ2θ1secθtanθsec2θdθ
where theta_1 = tan^-1sqrt3θ1=tan−1√3 and theta_2 = tan^-1sqrt35θ2=tan−1√35
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)∫θ2θ1sec3θtanθdθ=∫θ2θ1dθsinθcos2θ
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)
