Question

高数，不定积分，上一步是怎么到下一步的？

如图

Answer 1

1/(u^2+a^2)^2 = (1/a^2)[a^2/(u^2+a^2)^2]
= (1/a^2)[(a^2+u^2-u^2)/(u^2+a^2)^2]
= (1/a^2)[1/(u^2+a^2) - u^2/(u^2+a^2)^2]
I = ∫du/(u^2+a^2)^2 = (1/2a^2){2∫du/(u^2+a^2) - ∫2u^2du/(u^2+a^2)^2}
= (1/2a^2){2∫du/(u^2+a^2) - ∫ud(u^2+a^2)/(u^2+a^2)^2}
= (1/2a^2){2∫du/(u^2+a^2) + ∫ud[1/(u^2+a^2)]}
= (1/2a^2)[2∫du/(u^2+a^2) + u/(u^2+a^2) - ∫du/(u^2+a^2)]
= (1/2a^2)[u/(u^2+a^2) + ∫du/(u^2+a^2)]

Answer 2

积分公式！
∫u'dv'=u'v'-∫v'du'
这里，令 u'=1/[2a^2(u^2+a^2)], v'=u/2a^2

Answer 3

∫[1/(u²+a²)²]du=(1/a²)∫[a²/(u²+a²)²]du
=(1/a²)∫[(u²+a²-u²)/(u²+a²)²]du
=(1/a²)[∫[1/(u²+a²)]du-∫u²du/(u²+a²)²]
=(1/a²)[∫[1/(u²+a²)]du-(1/2)∫ud(u²+a²)/(u²+a²)²]
=(1/a²){∫[1/(u²+a²)]du+(1/2)∫ud[1/(u²+a²)]}
=(1/a²){∫[1/(u²+a²)]du+(1/2)[u/(u²+a²)-∫du/(u²+a²)]}
=(1/a²){(1/2)∫du/(u²+a²)+(1/2)[u/(u²+a²)]}
=(1/2a²)[u/(u²+a²)+∫du/(u²+a²)];