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

Answer: (B)

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

Answer 2

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:

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