①∫tan^10×sec^2xdx②∫[x/√(x^2-2)dx③∫[(2x-3)/(x^2-3x+8)]dx④∫(1/
①∫tan^10×sec^2xdx②∫[x/√(x^2-2)dx③∫[(2x-3)/(x^2-3x+8)]dx④∫(1/
∫tan^2xdx=∫(sec^2x-1)dx
∫sec^4x dx ∫sec^2x tan^2x dx
∫ sec^2 x dx
请解释高数例题:1、∫tan ^2 x sec xdx 2、∫1/x^2+4 dx 3、∫tanx dsec^(n-2)
∫secx/sec^2x-1 dx
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求积分∫(sec^2x/2+tan^2x)dx