The integral \( I=\int\displaystyle\frac{(1+\log x)^{3}}{x}\,dx \) equals:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{(1+\log x)^{4}}{4}+C \), Where \( C \) is an arbitrary constant
The derivative of \( 1+\log x \) is \( \frac{1}{x} \), which is already present, so let
\( \displaystyle t=1+\log x \)
\( \displaystyle dt=\frac{dx}{x} \)
Then
\( \displaystyle I=\int t^{3}\,dt \)
\( \displaystyle =\frac{t^{4}}{4}+C \)
\( \displaystyle =\frac{(1+\log x)^{4}}{4}+C \)