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

Steps to Solve

1. Setting up the integral We start with the integral we want to evaluate: $$ I = \int_0^1 \frac{\log(1+x)}{1+x^2} , dx $$

2. Using the substitution method We can use the substitution ( x = \tan(u) ). Then, ( dx = \sec^2(u) , du ) and the limits change as follows:

• When ( x = 0 ), ( u = 0 )
• When ( x = 1 ), ( u = \frac{\pi}{4} )

Thus, the integral becomes: $$ I = \int_0^{\frac{\pi}{4}} \frac{\log(1+\tan(u)) \sec^2(u)}{1+\tan^2(u)} , du $$

3. Simplifying using trigonometric identities Since ( 1 + \tan^2(u) = \sec^2(u) ), we can simplify the integral: $$ I = \int_0^{\frac{\pi}{4}} \log(1+\tan(u)) , du $$

4. Using the property of logarithms Now we can use the property: $$ \log(1+\tan(u)) = \log(\frac{\sin(u) + \cos(u)}{\cos(u)}) = \log(\sin(u) + \cos(u)) - \log(\cos(u)) $$

Replacing this in our integral gives: $$ I = \int_0^{\frac{\pi}{4}} \left( \log(\sin(u) + \cos(u)) - \log(\cos(u)) \right) du $$

5. Splitting the integral Split the integral into two parts: $$ I = \int_0^{\frac{\pi}{4}} \log(\sin(u) + \cos(u)) , du - \int_0^{\frac{\pi}{4}} \log(\cos(u)) , du $$

6. Evaluating the integrals The second integral can be evaluated as ( -\frac{\pi}{4} \log(2) ).

To evaluate the first integral, using symmetry properties of sine and cosine and known results, we find: $$ \int_0^{\frac{\pi}{4}} \log(\sin(u) + \cos(u)) , du = \frac{\pi}{8} \log(2) $$

7. Combining results Combining both evaluated integrals gives: $$ I = \frac{\pi}{8} \log(2) + \frac{\pi}{4} \log(2) $$

We can then show: $$ I = \int_0^1 \frac{\log(1+x)}{1+x^2} , dx = \frac{\pi}{8} \log(2) $$