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If x71+x42dx=14log1+x4+f(x)+C then 

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a
f(x)=1+x4
b
f(x)=11+x42
c
f(x)=11+x4
d
f(x)=tan−1⁡1+x4

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detailed solution

Correct option is C

Put  1+x4=t , so that ∫x71+x42dx=14∫(t−1)t2dt=14log⁡|t|+141t+C=14log⁡1+x4+11+x4+C∴f(x)=11+x4


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