Integral \( \mathrm{I}=\int\displaystyle\frac{e^{3\log x}-3x^4}{e^{2\log x}+x^2}\,dx \) (where \( x\gt 0 \)) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{1}{4}x^2-\frac{1}{2}x^3+C \), where \( C \) is an arbitrary constant
Since \( e^{k\log x}=e^{\log x^k}=x^k \) for x > 0:
\( \displaystyle e^{3\log x}=x^3 \)
\( \displaystyle e^{2\log x}=x^2 \)
So the integrand is
\( \displaystyle \frac{x^3-3x^4}{x^2+x^2} \)
\( \displaystyle =\frac{x^2(x-3x^2)}{2x^2} \)
\( \displaystyle =\frac{x}{2}-\frac{3x^2}{2} \)
\( \displaystyle \mathrm{I}=\int\left(\frac{x}{2}-\frac{3x^2}{2}\right)dx \)
\( \displaystyle =\frac{x^2}{4}-\frac{x^3}{2}+C \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{1}{2}x-\frac{3}{2}x^2+C \), where \( C \) is an arbitrary constant — Incorrect. This is the simplified integrand \( \frac{x}{2}-\frac{3x^2}{2} \) itself, not its integral.
Option (2) \( \displaystyle\frac{1}{4}x-\frac{1}{2}x^2+C \), where \( C \) is an arbitrary constant — Incorrect. The derivative of \( \frac{1}{4}x-\frac{1}{2}x^2 \) is \( \frac{1}{4}-x \), not \( \frac{x}{2}-\frac{3x^2}{2} \).
Option (3) \( \displaystyle\frac{1}{2}x^2-\frac{3}{4}x^3+C \), where \( C \) is an arbitrary constant — Incorrect. The derivative of \( \frac{1}{2}x^2-\frac{3}{4}x^3 \) is \( x-\frac{9}{4}x^2 \), not \( \frac{x}{2}-\frac{3x^2}{2} \).