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dxa2+x23/2 is equal to

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a
xa2+x21/2+C
b
xa2a2+x21/2+C
c
1a2a2+x21/2+C
d
None of these

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detailed solution

Correct option is B

I=∫dxa2+x23/2put , x=atan⁡θ⇒dx=asec2⁡θdθ∴ I=∫asec2⁡θa2+a2tan2⁡θ3/2dθ=∫asec2⁡θa3sec2⁡θ3/2dθ⇒ I=1a2∫dθsec⁡θ=1a2∫cos⁡θdθ=1a2sin⁡θ+C⇒ =xa2x2+a21/2+C


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