Question

Evaluate the integral: \[ \int \sqrt{x + \sqrt{x^2 + 2}} \, dx. \]

• A. \(\frac{2}{3} (x + \sqrt{x^2 + 2})^{3/2} - 2(x + \sqrt{x^2 + 2})^{1/2} + C\)
• B. \(\frac{1}{3} (x + \sqrt{x^2 + 2})^{3/2} - 2(x + \sqrt{x^2 + 2})^{1/2} + C\)
• C. \( (x + \sqrt{x^2 + 2})^{-3/2} - 2(x + \sqrt{x^2 + 2})^{1/2} + C\)
• D. \(\frac{(x+\sqrt{x^2+2})^2 - 6}{3\sqrt{x+\sqrt{x^2+2}}} + C\)

Hint:

Use substitution to simplify complicated square roots and reduce the integrand to a manageable form.

Answer 1

The Correct Option is D

By employing the substitution \( u = x + \sqrt{x^2 + 2} \), we proceed to calculate the integral:
\[int \sqrt{x + \sqrt{x^2 + 2}} \, dx\] \[\text{Let } \sqrt{x + \sqrt{x^2 + 2}} = t \Rightarrow x + \sqrt{x^2 + 2} = t^2\] \[\sqrt{x^2 + 2} = t^2 - x\] \[\Rightarrow x^2 + 2 = t^4 + x^2 - 2t^2x\] \[\Rightarrow x = \frac{t^4 - 2}{2t^2} \Rightarrow dx = \frac{t^4 + 2}{t^3} \, dt\] \[\int t \cdot \frac{t^4 + 2}{t^3} \, dt = \int \left(t^2 + \frac{2}{t^2}\right) \, dt = \frac{t^3}{3} - \frac{2}{t} + C\] \[= \frac{t^4 - 6}{3t} + C\] \[= \frac{(x + \sqrt{x^2 + 2})^2 - 6}{3\sqrt{x + \sqrt{x^2 + 2}}} + C\]