Q89 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics
If $f(x) = \dfrac{1}{\log x}$ and $g(x) = \dfrac{1}{(\log x)^2}$, then the value of $\displaystyle\int [f(x) - g(x)]\,dx$ is...
- A$(\log x)^2 + c$
- B$x\log x + c$
- Correct answerC$\dfrac{x}{\log x} + c$
- D$\dfrac{1}{\log x} + c$
Solution
Let $I = \int \left( \dfrac{1}{\log x} - \dfrac{1}{(\log x)^2} \right) dx$
Substitute $\log x = t \Rightarrow x = e^t \Rightarrow dx = e^t dt$
$I = \int e^t \left( \dfrac{1}{t} - \dfrac{1}{t^2} \right) dt$
Using the standard integral $\int e^t (h(t) + h'(t)) dt = e^t h(t) + c$
Here, $h(t) = \dfrac{1}{t}$ and $h'(t) = -\dfrac{1}{t^2}$
$I = e^t \left( \dfrac{1}{t} \right) + c$
Substituting back $t = \log x$ and $e^t = x$
$I = \dfrac{x}{\log x} + c$
Answer: $\dfrac{x}{\log x} + c$
Substitute $\log x = t \Rightarrow x = e^t \Rightarrow dx = e^t dt$
$I = \int e^t \left( \dfrac{1}{t} - \dfrac{1}{t^2} \right) dt$
Using the standard integral $\int e^t (h(t) + h'(t)) dt = e^t h(t) + c$
Here, $h(t) = \dfrac{1}{t}$ and $h'(t) = -\dfrac{1}{t^2}$
$I = e^t \left( \dfrac{1}{t} \right) + c$
Substituting back $t = \log x$ and $e^t = x$
$I = \dfrac{x}{\log x} + c$
Answer: $\dfrac{x}{\log x} + c$
Hence, the correct answer is option: 'C'.