Q89 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics
$\int \dfrac{\sqrt{x-2}}{x}dx = $
- Correct answerA$2\sqrt{x-2} - 2\sqrt{2}\tan^{-1}\left(\dfrac{\sqrt{x-2}}{\sqrt{2}}\right) + c$
- B$2\sqrt{x-2} + 2\sqrt{2}\tan^{-1}\left(\dfrac{\sqrt{x-2}}{\sqrt{2}}\right) + c$
- C$2\sqrt{x-2} - 2\sqrt{2}\tan^{-1}\left(\dfrac{\sqrt{x-2}}{2}\right) + c$
- D$2\sqrt{x-2} + 2\sqrt{2}\tan^{-1}\left(\dfrac{\sqrt{x-2}}{2}\right) + c$
Solution
Let $x - 2 = t^2$, which gives $dx = 2t dt$ and $x = t^2 + 2$.
Substituting these into the integral:
$I = \int \dfrac{t}{t^2 + 2} (2t dt) = 2 \int \dfrac{t^2}{t^2 + 2} dt$
$I = 2 \int \dfrac{t^2 + 2 - 2}{t^2 + 2} dt = 2 \int \left( 1 - \dfrac{2}{t^2 + 2} \right) dt$
$I = 2 \left( t - \dfrac{2}{\sqrt{2}} \tan^{-1} \left( \dfrac{t}{\sqrt{2}} \right) \right) + c$
$I = 2t - 2\sqrt{2} \tan^{-1} \left( \dfrac{t}{\sqrt{2}} \right) + c$
Substituting back $t = \sqrt{x-2}$:
$I = 2\sqrt{x-2} - 2\sqrt{2} \tan^{-1} \left( \dfrac{\sqrt{x-2}}{\sqrt{2}} \right) + c$
Answer: $2\sqrt{x-2} - 2\sqrt{2}\tan^{-1}\left(\dfrac{\sqrt{x-2}}{\sqrt{2}}\right) + c$
Substituting these into the integral:
$I = \int \dfrac{t}{t^2 + 2} (2t dt) = 2 \int \dfrac{t^2}{t^2 + 2} dt$
$I = 2 \int \dfrac{t^2 + 2 - 2}{t^2 + 2} dt = 2 \int \left( 1 - \dfrac{2}{t^2 + 2} \right) dt$
$I = 2 \left( t - \dfrac{2}{\sqrt{2}} \tan^{-1} \left( \dfrac{t}{\sqrt{2}} \right) \right) + c$
$I = 2t - 2\sqrt{2} \tan^{-1} \left( \dfrac{t}{\sqrt{2}} \right) + c$
Substituting back $t = \sqrt{x-2}$:
$I = 2\sqrt{x-2} - 2\sqrt{2} \tan^{-1} \left( \dfrac{\sqrt{x-2}}{\sqrt{2}} \right) + c$
Answer: $2\sqrt{x-2} - 2\sqrt{2}\tan^{-1}\left(\dfrac{\sqrt{x-2}}{\sqrt{2}}\right) + c$
Hence, the correct answer is option: 'A'.