Q94 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics
The value of the integral $\int\limits_{1/e}^{e}\dfrac{|\log x|}{x^2}dx$ is
- A$\dfrac{e^2 - 1}{2e}$
- B$\dfrac{2}{e}$
- Correct answerC$2 - \dfrac{2}{e}$
- D$1 - \dfrac{1}{e}$
Solution
Let $I = \int\limits_{1/e}^{e}\dfrac{|\log x|}{x^2}dx$
Substitute $x = e^t \Rightarrow dx = e^t dt$
When $x = 1/e$, $t = -1$ and when $x = e$, $t = 1$
$I = \int\limits_{-1}^{1} \dfrac{|t|}{e^{2t}} e^t dt = \int\limits_{-1}^{1} |t| e^{-t} dt$
Splitting the integral at $t = 0$:
$I = \int\limits_{-1}^{0} -t e^{-t} dt + \int\limits_{0}^{1} t e^{-t} dt$
Using integration by parts:
$\int -t e^{-t} dt = t e^{-t} + e^{-t}$
$\int t e^{-t} dt = -t e^{-t} - e^{-t}$
Evaluating the definite integrals:
$I = \left[ t e^{-t} + e^{-t} \right]_{-1}^{0} + \left[ -t e^{-t} - e^{-t} \right]_{0}^{1}$
$I = \left( (0 + e^0) - (-e^1 + e^1) \right) + \left( (-e^{-1} - e^{-1}) - (0 - e^0) \right)$
$I = (1 - 0) + \left( -\dfrac{2}{e} - (-1) \right)$
$I = 1 + 1 - \dfrac{2}{e} = 2 - \dfrac{2}{e}$
Answer: $2 - \dfrac{2}{e}$
Substitute $x = e^t \Rightarrow dx = e^t dt$
When $x = 1/e$, $t = -1$ and when $x = e$, $t = 1$
$I = \int\limits_{-1}^{1} \dfrac{|t|}{e^{2t}} e^t dt = \int\limits_{-1}^{1} |t| e^{-t} dt$
Splitting the integral at $t = 0$:
$I = \int\limits_{-1}^{0} -t e^{-t} dt + \int\limits_{0}^{1} t e^{-t} dt$
Using integration by parts:
$\int -t e^{-t} dt = t e^{-t} + e^{-t}$
$\int t e^{-t} dt = -t e^{-t} - e^{-t}$
Evaluating the definite integrals:
$I = \left[ t e^{-t} + e^{-t} \right]_{-1}^{0} + \left[ -t e^{-t} - e^{-t} \right]_{0}^{1}$
$I = \left( (0 + e^0) - (-e^1 + e^1) \right) + \left( (-e^{-1} - e^{-1}) - (0 - e^0) \right)$
$I = (1 - 0) + \left( -\dfrac{2}{e} - (-1) \right)$
$I = 1 + 1 - \dfrac{2}{e} = 2 - \dfrac{2}{e}$
Answer: $2 - \dfrac{2}{e}$
Hence, the correct answer is option: 'C'.