Q93 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics
The value of the definite integral $\int_0^{\log_e 5}\dfrac{e^x\sqrt{e^x-1}}{e^x+3}\,dx$ is equal to...
- A$5 + \pi$
- B$5 - \pi$
- C$4 + \pi$
- Correct answerD$4 - \pi$
Solution
Let $e^x - 1 = t^2$, which gives $e^x dx = 2t dt$.
When $x = 0$, $t = 0$.
When $x = \log_e 5$, $t = 2$.
The integral becomes:
$I = \int_0^2 \dfrac{t}{t^2 + 1 + 3} (2t dt) = \int_0^2 \dfrac{2t^2}{t^2 + 4} dt$
$I = 2 \int_0^2 \dfrac{t^2 + 4 - 4}{t^2 + 4} dt$
$I = 2 \int_0^2 \left( 1 - \dfrac{4}{t^2 + 4} \right) dt$
$I = 2 \left[ t - 4 \times \dfrac{1}{2} \tan^{-1} \left( \dfrac{t}{2} \right) \right]_0^2$
$I = 2 \left[ 2 - 2 \tan^{-1}(1) \right]$
$I = 2 \left( 2 - 2 \times \dfrac{\pi}{4} \right)$
$I = 4 - \pi$
Answer: $4 - \pi$
When $x = 0$, $t = 0$.
When $x = \log_e 5$, $t = 2$.
The integral becomes:
$I = \int_0^2 \dfrac{t}{t^2 + 1 + 3} (2t dt) = \int_0^2 \dfrac{2t^2}{t^2 + 4} dt$
$I = 2 \int_0^2 \dfrac{t^2 + 4 - 4}{t^2 + 4} dt$
$I = 2 \int_0^2 \left( 1 - \dfrac{4}{t^2 + 4} \right) dt$
$I = 2 \left[ t - 4 \times \dfrac{1}{2} \tan^{-1} \left( \dfrac{t}{2} \right) \right]_0^2$
$I = 2 \left[ 2 - 2 \tan^{-1}(1) \right]$
$I = 2 \left( 2 - 2 \times \dfrac{\pi}{4} \right)$
$I = 4 - \pi$
Answer: $4 - \pi$
Hence, the correct answer is option: 'D'.