Question

Evaluate the integral: $$ \int_0^1 \frac{\ln (1+x)}{1+x^2} \, dx $$

• A. \(\frac{\pi}{8} \ln 2\)
• B. \(\frac{\pi}{4} \ln 2\)
• C. \(\frac{\pi}{8} \ln (1+\sqrt{2})\)
• D. \(\frac{\pi}{4} \ln (1+\sqrt{2})\)

Hint:

For integrals involving logarithms and trigonometric denominators, use substitutions like \( x = \tan \theta \) and exploit symmetry to simplify.

Answer 1

The Correct Option is A

To evaluate the integral \(\int_0^1 \frac{\ln (1+x)}{1+x^2} \, dx\), we use the substitution \( x = \tan \theta \). 
Then, \( dx = \sec^2 \theta \, d\theta \), and the limits change as follows: when \( x = 0 \), \( \theta = 0 \); when \( x = 1 \), \( \theta = \frac{\pi}{4} \). Also, \( 1 + x = 1 + \tan \theta \), and \( 1 + x^2 = 1 + \tan^2 \theta = \sec^2 \theta \). The integral becomes: \[ \int_0^{\pi/4} \frac{\ln (1 + \tan \theta)}{\sec^2 \theta} \cdot \sec^2 \theta \, d\theta = \int_0^{\pi/4} \ln (1 + \tan \theta) \, d\theta \] Let \( I = \int_0^{\pi/4} \ln (1 + \tan \theta) \, d\theta \). Consider the substitution \( \phi = \frac{\pi}{4} - \theta \), so \( d\phi = -d\theta \), and the limits reverse: when \( \theta = 0 \), \( \phi = \frac{\pi}{4} \); when \( \theta = \frac{\pi}{4} \), \( \phi = 0 \). Thus: \[ I = \int_{\pi/4}^0 \ln \left(1 + \tan \left(\frac{\pi}{4} - \phi\right)\right) (-d\phi) = \int_0^{\pi/4} \ln \left(1 + \tan \left(\frac{\pi}{4} - \phi\right)\right) d\phi \] Using the identity \( \tan \left(\frac{\pi}{4} - \phi\right) = \frac{1 - \tan \phi}{1 + \tan \phi} \), we get: \[ 1 + \tan \left(\frac{\pi}{4} - \phi\right) = 1 + \frac{1 - \tan \phi}{1 + \tan \phi} = \frac{1 + \tan \phi + 1 - \tan \phi}{1 + \tan \phi} = \frac{2}{1 + \tan \phi} \] Thus: \[ \ln \left(1 + \tan \left(\frac{\pi}{4} - \phi\right)\right) = \ln \left(\frac{2}{1 + \tan \phi}\right) = \ln 2 - \ln (1 + \tan \phi) \] So: \[ I = \int_0^{\pi/4} \left( \ln 2 - \ln (1 + \tan \phi) \right) d\phi = \int_0^{\pi/4} \ln 2 \, d\phi - \int_0^{\pi/4} \ln (1 + \tan \phi) \, d\phi \] The second integral is \( I \), so: \[ I = \left[ \ln 2 \cdot \phi \right]_0^{\pi/4} - I = \ln 2 \cdot \frac{\pi}{4} - I \] Solving for \( I \): \[ 2I = \frac{\pi}{4} \ln 2 \implies I = \frac{\pi}{8} \ln 2 \] Thus, the value of the integral is: \[ \boxed{\frac{\pi}{8} \ln 2} \]