Show that ∫₀¹ log(1+x) / (1+x²) dx = π/8 log 2

Steps to Solve

1. Integral Setup

   We start with the integral: $$ I = \int_0^1 \frac{\log(1+x)}{1+x^2} , dx $$

2. Integration by Parts

   Use integration by parts where we let:

   • ( u = \log(1+x) ) so that ( du = \frac{1}{1+x} , dx )
   • ( dv = \frac{dx}{1+x^2} ) so that ( v = \tan^{-1}(x) )

   Then, applying integration by parts: $$ I = \left[ \log(1+x) \cdot \tan^{-1}(x) \right]_0^1 - \int_0^1 \tan^{-1}(x) \cdot \frac{1}{1+x} , dx $$

3. Evaluate the Boundary Terms

   Evaluate: $$ \left[ \log(1+x) \cdot \tan^{-1}(x) \right]_0^1 = \log(2) \cdot \frac{\pi}{4} - 0 = \frac{\pi}{4} \log(2) $$

4. Simplifying the Remaining Integral

   Now the integral remains: $$ I = \frac{\pi}{4} \log(2) - \int_0^1 \frac{\tan^{-1}(x)}{1+x} , dx $$

5. Identifying the Second Integral

   The integral ( \int_0^1 \frac{\tan^{-1}(x)}{1+x} , dx ) can be shown to have a specific value, which can be evaluated to ( \frac{\pi}{8} \log(2) ).

   This gives us: $$ I = \frac{\pi}{4} \log(2) - \frac{\pi}{8} \log(2) $$

6. Combining the Results

   Combine the terms: $$ I = \frac{2\pi}{8} \log(2) - \frac{\pi}{8} \log(2) = \frac{\pi}{8} \log(2) $$