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dx(a2+x2)32 =

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a
x(a2+x2)32+c
b
xa2(a2+x2)32+c
c
1a2(a2+x2)32+c
d
None of these

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detailed solution

Correct option is B

Put x=a tanθ ⇒dx=asec2θ dθ​∴I=∫asec2θa2+a2tan2θ32dθ​       =∫asec2θa3sec2θ32dθ​       =1a2∫dθsecθ​       =1a2∫cosθ dθ      ​=1a2sinθ+c​​       =xa2a2+x232+c


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