If \( \int\displaystyle e^{3\log x}(x^4+1)^{-1}\,dx=\displaystyle\frac{m}{n}\log(x^4+1)+C \), where \( C \) is an arbitrary constant, \( m \) and \( n \) are positive integers such that g.c.d (\( m,n \)) = 1, then (\( 4m^2+n^2 \) ) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 20
Since \( e^{3\log x}=e^{\log x^3}=x^3 \),
\( \displaystyle \int\frac{x^3}{x^4+1}\,dx=\frac{1}{4}\int\frac{4x^3}{x^4+1}\,dx \)
\( \displaystyle =\frac{1}{4}\log(x^4+1)+C \)
The numerator \( 4x^3 \) is the derivative of the denominator, which gives the log. So m = 1 and n = 4 (with g.c.d. 1).
\( 4m^2+n^2=4+16=20 \)