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B
c−1acos−1a|x|
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C
sin−1a|x|+c
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D
c+1asin−1a|x|
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Solution
The correct option is Dc−1asin−1a|x| Let I=∫dxx√x2−a2 Let x=1t ∴dx=−1t2dt ∴I=∫−dtt2.1t√(1t2)2−a2 =−1a∫dt(1a)2−t2=−1asin−1t+c =−1asin−1a|x|+c=c−1asin−1a|x|