∫[cosx+1/(1+x²)-1/√(1-x²)]dx=

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∫[cosx+1/(1+x²)-1/√(1-x²)]dx=∫[cosx+1/(1+x²)-1/√(1-x²)]dx=∫[cosx+1/(1+x²)-1

∫[cosx+1/(1+x²)-1/√(1-x²)]dx=
∫[cosx+1/(1+x²)-1/√(1-x²)]dx=

∫[cosx+1/(1+x²)-1/√(1-x²)]dx=
原积分=sinx+arctanx-arcsinx+C