Question

The value of the integral \[ \int_{-1}^{2} \log_e \left( x + \sqrt{x^2 + 1} \right) \, dx \] is:

• A. \( \sqrt{5} - \sqrt{2} + \log_e \left( \frac{9 + 4\sqrt{5}}{1 + \sqrt{2}} \right) \)
• B. \( \sqrt{2} - \sqrt{5} + \log_e \left( \frac{9 + 4\sqrt{5}}{1 + \sqrt{2}} \right) \)
• C. \( \sqrt{5} - \sqrt{2} + \log_e \left( \frac{7 + 4\sqrt{5}}{1 + \sqrt{2}} \right) \)
• D. \( \sqrt{2} - \sqrt{5} + \log_e \left( \frac{7 + 4\sqrt{5}}{1 + \sqrt{2}} \right) \)

Answer 1

The Correct Option is B

Let:

\[ f(x) = \log_e \left( x + \sqrt{x^2 + 1} \right) .\]

Use substitution to simplify the integral. Define \( u = x + \sqrt{x^2 + 1} \), so:

\[ du = \left( 1 + \frac{x}{\sqrt{x^2 + 1}} \right) dx = \frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}} dx = \frac{u}{\sqrt{x^2 + 1}} dx. \]

Squaring \( u \), we find:

\[ u^2 = x^2 + 1 + 2x \sqrt{x^2 + 1}. \]

Rearrange:

\[ x \sqrt{x^2 + 1} = \frac{u^2 - x^2 - 1}{2}. \]

From symmetry and the bounds \( x \in [-1, 2] \), evaluate \( u \) at \( x = -1 \) and \( x = 2 \):

At \( x = -1 \), \( u = -1 + \sqrt{2} \).

At \( x = 2 \), \( u = 2 + \sqrt{5} \).

Substitute back into the integral and compute:

\[ \int_{-1}^{2} \log_e \left( x + \sqrt{x^2 + 1} \right) dx = \sqrt{2} - \sqrt{5} + \log_e \left( \frac{9 + 4\sqrt{5}}{1 + \sqrt{2}} \right). \]

Final Answer:

\[ \boxed{\sqrt{2} - \sqrt{5} + \log_e \left( \frac{9 + 4\sqrt{5}}{1 + \sqrt{2}} \right)} \]