Question

The value of the integral \[ \int_{\log_2}^{-\log_2} e^x \left( \log \left( e^x + \sqrt{1 + e^{2x}} \right) \right) dx \] is equal to:

• A. \(\log \left( \frac{2 + \sqrt{5}}{\sqrt{5}} \right)\)
• B. \(\log \left( \frac{2 + \sqrt{5}}{\sqrt{5}} \right) / 2\)
• C. \(\log \left( \frac{2\sqrt{5}}{\sqrt{5}} \right)\)
• D. \(\log \left( \frac{2 + \sqrt{5}}{\sqrt{5}} \right) / 2\)

Hint:

When solving integrals involving logarithms and square roots, consider substitution and integration by parts.

Answer 1

The Correct Option is D

Let \[ I = \int_{\log_2}^{-\log_2} e^x \left( \log \left( e^x + \sqrt{1 + e^{2x}} \right) \right) dx \] Substitute \( e^x = t \), so that \( e^x dx = dt \). The limits also change accordingly: When \( x = \log_2 \), \( t = 2 \); and when \( x = -\log_2 \), \( t = 1/2 \).
Now, apply integration by parts: \[ I = \left[ \ln \left( t + \sqrt{t^2 + 1} \right) \right]_{\frac{1}{2}}^{2} \] This yields the result \[ \Rightarrow \log \left( \frac{2 + \sqrt{5}}{\sqrt{5}} \right) / 2 \]