Evaluate : `int_(0)^(1)(log(1+x))/(1+x^2)dx`

Answer

To evaluate the integral

\[ I = \int_0^1 \frac{\log(1+x)}{1+x^2} \, dx, \]

we will use the substitution \( x = \tan \theta \).

### Step 1: Substitution
Let \( x = \tan \theta \). Then, the differential \( dx = \sec^2 \theta \, d\theta \).

### Step 2: Change of Limits
When \( x = 0 \), \( \theta = \tan^{-1}(0) = 0 \).
When \( x = 1 \), \( \theta = \tan^{-1}(1) = \frac{\pi}{4} \).

Thus, the limits change from \( 0 \) to \( \frac{\pi}{4} \).

### Step 3: Rewrite the Integral
Now, we rewrite the integral:

\[ I = \int_0^{\frac{\pi}{4}} \frac{\log(1+\tan \theta)}{1+\tan^2 \theta} \sec^2 \theta \, d\theta. \]

Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \), we can simplify:

\[ I = \int_0^{\frac{\pi}{4}} \log(1+\tan \theta) \, d\theta. \]

### Step 4: Symmetry in the Integral
Next, we will use the property of the integral. We can express \( I \) in another form by substituting \( \theta = \frac{\pi}{4} - t \):

\[ I = \int_0^{\frac{\pi}{4}} \log(1+\tan(\frac{\pi}{4}-t)) \, dt. \]

Using the identity \( \tan(\frac{\pi}{4}-t) = \frac{1 - \tan t}{1 + \tan t} \):

\[ I = \int_0^{\frac{\pi}{4}} \log\left(1 + \frac{1 - \tan t}{1 + \tan t}\right) \, dt. \]

### Step 5: Simplifying the Logarithm
The expression simplifies to:

\[ 1 + \frac{1 - \tan t}{1 + \tan t} = \frac{(1 + \tan t) + (1 - \tan t)}{1 + \tan t} = \frac{2}{1 + \tan t}. \]

Thus, we have:

\[ I = \int_0^{\frac{\pi}{4}} \log\left(\frac{2}{1+\tan t}\right) \, dt = \int_0^{\frac{\pi}{4}} \log(2) \, dt - \int_0^{\frac{\pi}{4}} \log(1+\tan t) \, dt. \]

### Step 6: Combining the Integrals
This gives us:

\[ I = \frac{\pi}{4} \log(2) - I. \]

### Step 7: Solving for \( I \)
Now, we can solve for \( I \):

\[ 2I = \frac{\pi}{4} \log(2) \implies I = \frac{\pi}{8} \log(2). \]

### Final Result
Thus, the value of the integral is:

\[ \boxed{\frac{\pi}{8} \log(2)}. \]