不定积分求解

2024-11-25 04:16:57
推荐回答(1个)
回答1:

∫4/(1-2x)²dx
=-2∫1/(1-2x)²)d(1-2x)
=2∫d1/(1-2x)
=2/(1-2x) + C (C是常数)

∫x(x²+3)^(1/2)dx
=1/2*∫(x²+3)^(1/2)d(x²+3)
=1/2*2/3*∫d(x²+3)^(3/2)
=[(x²+3)^(3/2)]/3 + C