积分计算：∫(x²/√(1+x²))dx

令u=1+x²，则du=2xdx，可得：

∫(x²/√(1+x²))dx = 1/2 ∫(u-1)/√u du

= 1/2 ∫(u^(1/2) - u^(-1/2))du

= 1/2 (2u^(3/2)/3 - 2u^(1/2)) + C

= 1/3 (1+x²)^(3/2) - x√(1+x²) + C

所以∫(x²/√(1+x²))dx = 1/3 (1+x²)^(3/2) - x√(1+x²) + C。