Question

∫x²/（（a²+x²）½）dx 的不定积分怎么求?

Answer 1

∫(x²+a²-a²)/（（a²+x²）½）dx
=∫（a²+x²）½dx-a²∫1/（（a²+x²）½）dx
令x=atanu，则（a²+x²）½=asecu，dx=a(secu)²du
原式=∫
asecu*a(secu)²du-a²∫
1/asecu*a(secu)²du
=a²∫
(secu)^3du-a²∫
secudu
=a²∫
(secu)^3du-a²ln|secu+tanu|
(1)
下面计算
∫
(secu)^3du
=∫
secudtanu
=secu*tanu-∫
(tanu)²secudu
=secu*tanu-∫
[(secu)²-1]secudu
=secu*tanu-∫
(secu)^3du+∫
secudu
将-∫
(secu)^3du移动左边与左边合并后除去系数得
∫
(secu)^3du=1/2secu*tanu+1/2∫
secudu+C1
=1/2secu*tanu+1/2ln|secu+tanu|+C1
代回（1）得：
原式=a²/2*secu*tanu-a²/2*ln|secu+tanu|+C2
代回原变量
原式=a²/2*
[√(a²+x²)/a]*(x/a)-a²/2*ln|√(a²+x²)/a+x/a|+C2
=1/2*
[x√(a²+x²)]-a²/2*ln|√(a²+x²)+x|+C