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 If 2x14xdx=Ksin12x+C, then K is equal to 

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a
log⁡2
b
12log⁡2
c
12
d
1log⁡2

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detailed solution

Correct option is D

Let I=∫2x1−4xdx=1log⁡2∫11−t2dt Putting 2x=t,2xlog⁡2dx=dtI=1log⁡2sin−1⁡t1+C=1log⁡2sin−1⁡2x+C∴ K=1log⁡2


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