lines x=0 and 1, y=..., around x-axis: (i'll make { the interal sign..., and put between 1 and 0 in front)
around x-axis so V = {(y squared)dx
y squared = 2x/(x2 +1) -> squaring y removes the root over x
So V = pi (1,0){(y2) dx
= pi (1,0){ 2x/(x2 +1) dx -> integrate - it's a log so...
= pi [ln(x2 +1)](1,0) -> sub in (1,0) values
= pi [ln(2) - ln(1)] -> since ln1 = 0
= ln2pi units cubed
It's a little awkward because of the limited text availability, but hope that makes sense.