The integral \( I=\int\displaystyle\frac{(1+\log_e x)^3}{x}\,dx \) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{(1+\log_e x)^4}{4}+C,\ where\ C\ is\ an\ arbitrary\ constant \)
The derivative of \( 1+\log_e x \) is \( \frac{1}{x} \), which is already in the integrand, so a substitution works. Let
\( t=1+\log_e x \)
\( \displaystyle dt=\frac{dx}{x} \)
Then
\( \displaystyle I=\int t^3\,dt \)
\( \displaystyle =\frac{t^4}{4}+C \)
\( \displaystyle =\frac{(1+\log_e x)^4}{4}+C \)