∫(1/a+bx^2) •dx= 1/√ab •arc tan( x√ab)/a+C (a、b同号)
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∫(1/a+bx^2) •dx= 1/√ab •arc tan( x√ab)/a+C (a、b同号)
∫(1/a+bx^2) •dx= 1/√ab •arc tan( x√ab)/a+C (a、b同号)
∫(1/a+bx^2) •dx= 1/√ab •arc tan( x√ab)/a+C (a、b同号)
换元,令 u = √(b/a) x ,dx = √(a/b) du
I = (1/a) ∫ 1/[1+(b/a)x²] dx
= (1/a) ∫ 1/[1 + u²] √(a/b) du
= 1/√ab arctanu + C
= ﹍﹍