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Question : 76 of 160
Marks:
+1,
-0
Solution:
Consider
Take log on both sides
logA=[log(1+)+log(1+) .....+log(1+)] logA=log(1+) =log (1+x2)dx Using integration by parts
logA=log(1+x2)∫1dx −∫(log(1+x2))∫1dx =xlog(1+x2)−∫xdx =xlog(1+x2)−2∫dx =[xlog(1+x2)−2x+2tan−1(x)]01 logA=[log2−2+2tan−1(1)] A=elog2−2+2()=2e
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