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Q.

If ∫x71+x42dx=14log⁡1+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

answer is C.

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Detailed Solution

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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