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The value of limx0xtan1x2dxx2+1 is

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a
π/4
b
π2/2
c
π2/4
d
π

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detailed solution

Correct option is C

limx→∞ ∫0x tan−1⁡x2dxx2+1=limx→∞ tan−1⁡x2x2+1x                                                          (L’ Hôpital Rule)=limx→∞ tan−1⁡x21+1x2=π22=π24


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