Question

\(\int \frac{\sqrt{x-2}}{x}dx =\)

• A. \(2\sqrt{x-2}-2\sqrt{2}tan^{-1}(\frac{\sqrt{x-2}}{\sqrt{2}})+c\)
• B. \(2\sqrt{x-2}+2\sqrt{2}tan^{-1}(\frac{\sqrt{x-2}}{\sqrt{2}})+c\)
• C. \(2\sqrt{x-2}-2\sqrt{2}tan^{-1}(\frac{\sqrt{x-2}}{2})+c\)
• D. \(2\sqrt{x-2}+2\sqrt{2}tan^{-1}(\frac{\sqrt{x-2}}{2})+c\)

Hint:

Let t = sqrt(x - 2) and split t^2 / (t^2 + 2).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Substitution removes the square root. Let \(t=\sqrt{x-2}\).

Step 2: Substitute:
Then \(x=t^2+2\) and \(dx=2t\,dt\).
\[ \int\frac{\sqrt{x-2}}{x}dx=\int\frac{t\cdot2t}{t^2+2}dt=2\int\frac{t^2}{t^2+2}dt \]

Step 3: Split the fraction:
\[ 2\int\left(1-\frac{2}{t^2+2}\right)dt=2t-4\int\frac{dt}{t^2+2} \]

Step 4: Integrate:
Using \(\int\dfrac{dt}{t^2+a^2}=\dfrac1a\tan^{-1}\dfrac ta\) with \(a=\sqrt2\):
\[ 2t-\frac{4}{\sqrt2}\tan^{-1}\frac t{\sqrt2}=2t-2\sqrt2\tan^{-1}\frac t{\sqrt2} \]

Step 5: Back-substitute:
\[ 2\sqrt{x-2}-2\sqrt2\tan^{-1}\frac{\sqrt{x-2}}{\sqrt2}+c \]
This is option (A). Options (B) and (D) carry a plus sign, and (C) and (D) divide by 2 instead of \(\sqrt2\) inside the tangent inverse.

Final Answer:
The result matches option (A). \[ \boxed{2\sqrt{x-2}-2\sqrt2\tan^{-1}\frac{\sqrt{x-2}}{\sqrt2}+c} \]