Q93 / 150 · MHT CET 2026 (11 April Shift 2) · Mathematics
If $\int_{2}^{e}\left[\dfrac{1}{\log x} - \dfrac{1}{(\log x)^2}\right] dx = a + \dfrac{b}{\log 2}$, then values of $a$ and $b$ are
- A$a = e, b = 2$
- B$a = -e, b = 2$
- Correct answerC$a = e, b = -2$
- D$a = -e, b = -2$
Solution
Let $I = \int_{2}^{e}\left[\dfrac{1}{\log x} - \dfrac{1}{(\log x)^2}\right] dx$
Using integration by parts on the first term:
$\int \dfrac{1}{\log x} dx = \dfrac{x}{\log x} - \int x \left( -\dfrac{1}{(\log x)^2} \right) \dfrac{1}{x} dx = \dfrac{x}{\log x} + \int \dfrac{1}{(\log x)^2} dx$
Therefore, $\int \left[ \dfrac{1}{\log x} - \dfrac{1}{(\log x)^2} \right] dx = \dfrac{x}{\log x}$
Evaluating the definite integral from $2$ to $e$:
$I = \left[ \dfrac{x}{\log x} \right]_{2}^{e} = \dfrac{e}{\log e} - \dfrac{2}{\log 2} = e - \dfrac{2}{\log 2}$
Comparing this with $a + \dfrac{b}{\log 2}$, we get $a = e$ and $b = -2$.
Answer: $a = e, b = -2$
Using integration by parts on the first term:
$\int \dfrac{1}{\log x} dx = \dfrac{x}{\log x} - \int x \left( -\dfrac{1}{(\log x)^2} \right) \dfrac{1}{x} dx = \dfrac{x}{\log x} + \int \dfrac{1}{(\log x)^2} dx$
Therefore, $\int \left[ \dfrac{1}{\log x} - \dfrac{1}{(\log x)^2} \right] dx = \dfrac{x}{\log x}$
Evaluating the definite integral from $2$ to $e$:
$I = \left[ \dfrac{x}{\log x} \right]_{2}^{e} = \dfrac{e}{\log e} - \dfrac{2}{\log 2} = e - \dfrac{2}{\log 2}$
Comparing this with $a + \dfrac{b}{\log 2}$, we get $a = e$ and $b = -2$.
Answer: $a = e, b = -2$
Hence, the correct answer is option: 'C'.