Q76 / 150 · MHT CET 2026 (16 April Shift 1) · Mathematics
$\displaystyle\int_0^{\log 5} \dfrac{e^x\sqrt{e^x - 1}}{e^x + 3}\,dx =$
- A$2 - \dfrac{\pi}{4}$
- B$4 - \dfrac{\pi}{4}$
- C$2 - \pi$
- Correct answerD$4 - \pi$
Solution
Let $e^x - 1 = t^2$. Differentiating both sides, we get $e^x dx = 2t dt$.
When $x = 0$, $t^2 = e^0 - 1 = 0 \Rightarrow t = 0$.
When $x = \log 5$, $t^2 = e^{\log 5} - 1 = 5 - 1 = 4 \Rightarrow t = 2$.
The denominator becomes $e^x + 3 = (t^2 + 1) + 3 = t^2 + 4$.
Substituting these into the integral, we get:
$\displaystyle\int_0^{\log 5} \dfrac{\sqrt{e^x - 1}}{e^x + 3} (e^x dx) = \int_0^2 \dfrac{t}{t^2 + 4} (2t dt)$
$= \displaystyle\int_0^2 \dfrac{2t^2}{t^2 + 4} dt$
$= 2 \displaystyle\int_0^2 \dfrac{t^2 + 4 - 4}{t^2 + 4} dt$
$= 2 \displaystyle\int_0^2 \left( 1 - \dfrac{4}{t^2 + 4} \right) dt$
$= 2 \left[ t - 4 \left( \dfrac{1}{2} \tan^{-1} \left( \dfrac{t}{2} \right) \right) \right]_0^2$
$= 2 \left[ t - 2 \tan^{-1} \left( \dfrac{t}{2} \right) \right]_0^2$
$= 2 \left[ \left( 2 - 2 \tan^{-1}(1) \right) - (0 - 0) \right]$
$= 2 \left( 2 - 2 \left( \dfrac{\pi}{4} \right) \right)$
$= 4 - \pi$
Answer: $4 - \pi$
When $x = 0$, $t^2 = e^0 - 1 = 0 \Rightarrow t = 0$.
When $x = \log 5$, $t^2 = e^{\log 5} - 1 = 5 - 1 = 4 \Rightarrow t = 2$.
The denominator becomes $e^x + 3 = (t^2 + 1) + 3 = t^2 + 4$.
Substituting these into the integral, we get:
$\displaystyle\int_0^{\log 5} \dfrac{\sqrt{e^x - 1}}{e^x + 3} (e^x dx) = \int_0^2 \dfrac{t}{t^2 + 4} (2t dt)$
$= \displaystyle\int_0^2 \dfrac{2t^2}{t^2 + 4} dt$
$= 2 \displaystyle\int_0^2 \dfrac{t^2 + 4 - 4}{t^2 + 4} dt$
$= 2 \displaystyle\int_0^2 \left( 1 - \dfrac{4}{t^2 + 4} \right) dt$
$= 2 \left[ t - 4 \left( \dfrac{1}{2} \tan^{-1} \left( \dfrac{t}{2} \right) \right) \right]_0^2$
$= 2 \left[ t - 2 \tan^{-1} \left( \dfrac{t}{2} \right) \right]_0^2$
$= 2 \left[ \left( 2 - 2 \tan^{-1}(1) \right) - (0 - 0) \right]$
$= 2 \left( 2 - 2 \left( \dfrac{\pi}{4} \right) \right)$
$= 4 - \pi$
Answer: $4 - \pi$
Hence, the correct answer is option: 'D'.