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

The integral \(\int_{-1}^{2} \log_e \left( x + \sqrt{x^2 + 1} \right) \, dx\) is solved using logarithmic properties and integration by parts. The steps are as follows:

1. \(\text{Let } x = \sinh t. \text{ Then, } dx = \cosh t \, dt \text{ and } \sqrt{x^2 + 1} = \cosh t.\)

Using hyperbolic identities:

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

1. \(\text{Note that } \sinh t + \cosh t = e^t. \text{ The integral transforms to:}\)

The limits are converted from \(x\) to \(t\):

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

The integral is evaluated:

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

The evaluated boundaries are calculated:

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

Values are computed using \(\text{arcsinh}(x) = \log_e(x + \sqrt{x^2 + 1})\):

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

Substituting these values yields:

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

Final computation by substituting values:

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

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