For \(x \gt 0\), \(\int\displaystyle\frac{e^{3\log x}}{x^4 + 1}\,dx\) is equal to (where C is a constant of integration)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{4}\log(x^4 + 1) + C\)
Since \(e^{\log a} = a\),
\(e^{3\log x} = e^{\log x^3}\)
\(= x^3\)
So the integral is \(\int \frac{x^3}{x^4 + 1}\,dx\). Let \(t = x^4 + 1\). Then \(dt = 4x^3\,dx\).
\( \displaystyle \int \frac{x^3}{x^4 + 1}\,dx = \frac{1}{4}\int \frac{dt}{t} \)
\( \displaystyle = \frac{1}{4}\log(x^4 + 1) + C \)
No modulus is needed because \(x^4 + 1 \gt 0\).