\( \int\left\{\displaystyle\frac{1}{\log x}-\displaystyle\frac{1}{(\log x)^2}\right\}dx \) is equal to:(where \( c \) is an arbitrary constant)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{x}{\log x}+c \)
Integrating the first term by parts, with \( \frac{1}{\log x} \) as the first function and 1 as the second,
\( \displaystyle \frac{d}{dx}\left(\frac{1}{\log x}\right)=-\frac{1}{(\log x)^2}\cdot\frac{1}{x} \)
\( \displaystyle \int\frac{1}{\log x}\,dx=\frac{x}{\log x}-\int x\cdot\left(-\frac{1}{x(\log x)^2}\right)dx \)
\( \displaystyle =\frac{x}{\log x}+\int\frac{1}{(\log x)^2}\,dx \)
Substituting in the given integral, the two integrals of \( \frac{1}{(\log x)^2} \) cancel:
\( \displaystyle \int\left\{\frac{1}{\log x}-\frac{1}{(\log x)^2}\right\}dx=\frac{x}{\log x}+c \)
Shortcut: Differentiate Option (3) with the quotient rule:
\( \displaystyle \frac{d}{dx}\left(\frac{x}{\log x}\right)=\frac{\log x-x\cdot\frac{1}{x}}{(\log x)^2} \)
\( \displaystyle =\frac{1}{\log x}-\frac{1}{(\log x)^2} \)
Tip: When an integrand has the form \( f(x)+xf'(x) \), the answer is \( xf(x)+c \), since that is the derivative of \( xf(x) \). Here \( f(x)=\frac{1}{\log x} \).