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If ∫e3log⁡x(x4+1)−1dx=mnlog⁡(x4+1)+C\int e^{3\log x}(x^4+1)^{-1}dx = \frac{m}{n}\log(x^4+1)+C, where C is an arbitrary constant, mm and nn are positive integers such that g.c.d (m,n)=1(m,n)=1, then (4m2+n2)(4m^2+n^2) is equal to

Solution

✅ Correct Option: 1

e3log⁡x=x3e^{3\log x} = x^3, so the integral is ∫x3x4+1dx=14log⁡(x4+1)+C\int\frac{x^3}{x^4+1}dx = \frac{1}{4}\log(x^4+1)+C.

Thus m=1m=1, n=4n=4 (coprime), and 4m2+n2=4+16=204m^2+n^2 = 4+16 = 20.

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