Q90 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics
$\int\dfrac{\log x}{(1 + \log x)^2}dx =$
- A$\dfrac{1}{1 + \log x} + c$
- B$-\dfrac{1}{1 + \log x} + c$
- Correct answerC$\dfrac{x}{1 + \log x} + c$
- D$-\dfrac{x}{1 + \log x} + c$
Solution
Let $\log x = t \Rightarrow x = e^t \Rightarrow dx = e^t dt$
Substituting in the integral:
$I = \int \dfrac{t}{(1 + t)^2} e^t dt$
$I = \int e^t \left( \dfrac{t + 1 - 1}{(1 + t)^2} \right) dt$
$I = \int e^t \left( \dfrac{1}{1 + t} - \dfrac{1}{(1 + t)^2} \right) dt$
This is of the form $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$, where $f(t) = \dfrac{1}{1 + t}$.
$I = \dfrac{e^t}{1 + t} + c$
Substituting $t = \log x$ and $e^t = x$:
$I = \dfrac{x}{1 + \log x} + c$
Answer: $\dfrac{x}{1 + \log x} + c$
Substituting in the integral:
$I = \int \dfrac{t}{(1 + t)^2} e^t dt$
$I = \int e^t \left( \dfrac{t + 1 - 1}{(1 + t)^2} \right) dt$
$I = \int e^t \left( \dfrac{1}{1 + t} - \dfrac{1}{(1 + t)^2} \right) dt$
This is of the form $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$, where $f(t) = \dfrac{1}{1 + t}$.
$I = \dfrac{e^t}{1 + t} + c$
Substituting $t = \log x$ and $e^t = x$:
$I = \dfrac{x}{1 + \log x} + c$
Answer: $\dfrac{x}{1 + \log x} + c$
Hence, the correct answer is option: 'C'.