积分计算：∫(x²/√(1+x²))dx 的详细步骤

令 u = 1 + x²，那么 du = 2xdx，即 xdx = du/2

∫(x²/√(1+x²))dx = ∫(u-1)/2√udu = (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)-(1/2)√(1+x²)+C，其中 C 为常数。