Q93 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics
$\int_0^{\log 10}\left[\dfrac{e^x \sqrt{e^x - 1}}{e^x + 8}\right]dx = $
- A$\dfrac{3}{2}(4 + \pi)$
- Correct answerB$\dfrac{3}{2}(4 - \pi)$
- C$\dfrac{3}{4}(4 - \pi)$
- D$\dfrac{3}{4}(4 + \pi)$
Solution
Let $e^x - 1 = t^2$, then $e^x dx = 2t dt$.
When $x = 0$, $t = 0$.
When $x = \log 10$, $t = 3$.
Substituting these into the integral, we get:
$I = \int_0^3 \dfrac{t}{t^2 + 1 + 8} (2t dt)$
$I = 2 \int_0^3 \dfrac{t^2}{t^2 + 9} dt$
$I = 2 \int_0^3 \left( 1 - \dfrac{9}{t^2 + 9} \right) dt$
$I = 2 \left[ t - 3 \tan^{-1} \left( \dfrac{t}{3} \right) \right]_0^3$
$I = 2 \left[ 3 - 3 \tan^{-1}(1) \right]$
$I = 2 \left( 3 - \dfrac{3\pi}{4} \right)$
$I = \dfrac{3}{2}(4 - \pi)$
Answer: $\dfrac{3}{2}(4 - \pi)$
When $x = 0$, $t = 0$.
When $x = \log 10$, $t = 3$.
Substituting these into the integral, we get:
$I = \int_0^3 \dfrac{t}{t^2 + 1 + 8} (2t dt)$
$I = 2 \int_0^3 \dfrac{t^2}{t^2 + 9} dt$
$I = 2 \int_0^3 \left( 1 - \dfrac{9}{t^2 + 9} \right) dt$
$I = 2 \left[ t - 3 \tan^{-1} \left( \dfrac{t}{3} \right) \right]_0^3$
$I = 2 \left[ 3 - 3 \tan^{-1}(1) \right]$
$I = 2 \left( 3 - \dfrac{3\pi}{4} \right)$
$I = \dfrac{3}{2}(4 - \pi)$
Answer: $\dfrac{3}{2}(4 - \pi)$
Hence, the correct answer is option: 'B'.