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B
−12log|tanxtanx+2|+c
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C
13log|tanxtanx+2|+c
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D
−13log|tanxtanx+2|+c
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Solution
The correct option is A12log|tanxtanx+2|+c ∫cosec2x1+2cotxdx=−∫dcotx1+2cotx =−12log∣1+2cotx∣+c =−12log∣1+2tanx∣+c =−12log∣tanx+2tanx∣+c =12log∣tanxtanx+2∣+c