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

To solve the integral \(\int_{-1}^{2} \log_e \left( x + \sqrt{x^2 + 1} \right) \, dx\), we can use the properties of logarithms and integration by parts. Let's break down the steps:

1. \(\text{Consider the substitution } x = \sinh t. \text{ Then, } dx = \cosh t \, dt. \text{ Thus, } \sqrt{x^2 + 1} = \cosh t.\)

By using the hyperbolic identities:

• \(x = \sinh t \Rightarrow \sqrt{x^2 + 1} = \cosh t.\)

1. \(\text{Note: } \sinh t + \cosh t = e^t. \text{ Therefore, the integral becomes:}\)

Change the limits from \(x\) to \(t\) conditions:

• \(x = -1 \Rightarrow \sinh t = -1 \Rightarrow t = \text{arcsinh}(-1)\)
• \(x = 2 \Rightarrow \sinh t = 2 \Rightarrow t = \text{arcsinh}(2)\)

Evaluate the integral:

1. \(\int t \, dt = \frac{t^2}{2} \bigg|_{\text{arcsinh}(-1)}^{\text{arcsinh}(2)}.\)

Calculate the evaluated boundaries:

1. \(\frac{1}{2} \left( (\text{arcsinh}(2))^2 - (\text{arcsinh}(-1))^2 \right).\)

Compute the values using the property \(\text{arcsinh}(x) = \log_e(x + \sqrt{x^2 + 1})\):

1. \(\text{arcsinh}(2) = \log_e(2 + \sqrt{5}),\ \text{arcsinh}(-1) = \log_e(-1 + \sqrt{2}).\)

Thus:

1. \(\frac{1}{2} \left( (\log_e (2 + \sqrt{5}))^2 - (\log_e (-1 + \sqrt{2}))^2 \right).\)

Put the values back to finalize the computation:

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

Thus, the value of the integral is \(\sqrt{2} - \sqrt{5} + \log_e \left( \frac{9 + 4\sqrt{5}}{1 + \sqrt{2}} \right).\)

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:

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