Verified answer

Let u = x^2 and du = 2xdx.

∫ x(1 + x^2)^(1/2) dx =

1/2 ∫ 2x(1 + x^2)^(1/2) dx =

1/2 ∫ (1 + u)^(1/2) du =

(1/2) 2(1 + u)^(3/2)/3 + c =

(1 + u)^(3/2)/3 + c =

(1 + x^2)^(3/2)/3 + c

put (1+x^2)=t

therefore 2x dx = dt

equation becomes

sqrt( t ) dt/2

intergrating we get t^(3/2) *4/3 +C

final answer 4(1+x^2) / 3 + C