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log(x+1+x2)1+x2dx

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a
12log(x+1+x2)2+c
b
log(x+1+x2)2+c
c
log(x+1+x2)+c
d
None of these

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detailed solution

Correct option is A

Let logx+1+x2 =t​Differentiating, we get​1x+1+x21+2x21+x2dx=dt​⇒11+x2dx=dt​∴∫logx+1+x21+x2dx=∫t dt​                                                   =t22+c​                                                                                            =12logx+1+x22+c


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