Integrate wrt y √(4y2+1). Let 2y=tanθ, then 2dy=sec2θdθ; 4y2=tan2θ, 4y2+1=sec2θ.
∫√(4y2+1)dy=½∫sec3θdθ=J/2 where J=∫sec3θdθ.
Let u=secθ, then du=secθtanθdθ; let dv=sec2θdθ, then v=tanθ.
Integrating by parts:
J=∫udv=secθtanθ-∫secθtan2θdθ=secθtanθ-∫secθ(sec2θ-1)dθ=secθtanθ-J+∫secθdθ.
2J=secθtanθ+∫secθdθ.
∫secθdθ=∫secθ[(secθ+tanθ)/(secθ+tanθ)]dθ=
∫[(sec2θ+secθtanθ)/(tanθ+secθ)]dθ=ln|tanθ+secθ|.
So J=½(secθtanθ+ln|tanθ+secθ|).
When y=0, tanθ=θ=0, secθ=1; when y=1, tanθ=2; secθ=√5.
Therefore ½J=¼(secθtanθ+ln|tanθ+secθ|. Applying the limits:
¼(2√5+ln(2+√5))=1.4789 approx.