3不定积分，过程，在线采纳，谢谢

Question

如题所述

Answer 1

用分部积分法, u = ln[x + √(1+x²)], dv = dx
du = {1 + 2x/[2√(1+x²)]}dx/[x + √(1+x²)] = dx/√(1+x²), v = x
原积分 = uv - ∫vdu = xln[x + √(1+x²)] - ∫xdx/√(1+x²)
= xln[x + √(1+x²)] - (1/2)∫d(1+x²)/√(1+x²)
= xln[x + √(1+x²)] - (1/2)*[1/(-1/2 + 1)]*(1+x²)^(-1/2 + 1)
= xln(x + √(1+x²)) - √(1+x²) + C