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∫tanx/(sin²;x + 1)dx=∫sinx/[cosx(1 - cos²;x + 1)] dx=∫sinx/[cosx(2 - cos²;x)] dx=∫1/[cosx(cos²;x - 2)] d(cosx)=(1/2)∫[cos²;x -(cos²;x - 2)]/[cosx(cos²;x - 2…
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