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Q.

If ∫2x1−4xdx=Ksin−1⁡2x+C, then K is equal to

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a

log⁡2

b

12log⁡2

c

12

d

1log⁡2

answer is D.

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Detailed Solution

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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If ∫2x1−4xdx=Ksin−1⁡2x+C, then K is equal to