`int_(0)^(1) (log (1+x))/(1+ x^(2)) dx` is equal to

Options

A. `(pi)/(8)`
B. `(pi)/(8) log 2`
C. log 2
D. None of these

Answer

To solve the integral

\[ I = \int_{0}^{1} \frac{\log(1+x)}{1+x^2} \, dx, \]

we can use a substitution and properties of definite integrals. Let's go through the steps:

### Step 1: Substitution
We will use the substitution \( x = \tan(\theta) \). Then, we have:

\[ dx = \sec^2(\theta) \, d\theta. \]

The limits change as follows:
- When \( x = 0 \), \( \theta = 0 \).
- When \( x = 1 \), \( \theta = \frac{\pi}{4} \).

### Step 2: Change the integral
Now, we can rewrite the integral in terms of \( \theta \):

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\log(1+\tan(\theta))}{1+\tan^2(\theta)} \sec^2(\theta) \, d\theta. \]

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

\[ I = \int_{0}^{\frac{\pi}{4}} \log(1+\tan(\theta)) \, d\theta. \]

### Step 3: Define a new integral
Let’s denote this integral as:

\[ I = \int_{0}^{\frac{\pi}{4}} \log(1+\tan(\theta)) \, d\theta. \]

### Step 4: Use symmetry property
We can use the property of integrals that states:

\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx. \]

In our case, we can set \( a = \frac{\pi}{4} \) and \( f(x) = \log(1+\tan(x)) \). Then we have:

\[ I = \int_{0}^{\frac{\pi}{4}} \log(1+\tan(\frac{\pi}{4} - x)) \, dx. \]

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

\[ I = \int_{0}^{\frac{\pi}{4}} \log\left(1 + \frac{1 - \tan(x)}{1 + \tan(x)}\right) \, dx. \]

### Step 5: Simplify the logarithm
We simplify the logarithm:

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

Thus,

\[ I = \int_{0}^{\frac{\pi}{4}} \log\left(\frac{2}{1+\tan(x)}\right) \, dx. \]

### Step 6: Break the logarithm
Using the property of logarithms:

\[ I = \int_{0}^{\frac{\pi}{4}} \log(2) \, dx - \int_{0}^{\frac{\pi}{4}} \log(1+\tan(x)) \, dx. \]

The first integral evaluates to:

\[ \log(2) \cdot \frac{\pi}{4}. \]

### Step 7: Combine the results
Now we have:

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

Adding \( I \) to both sides gives:

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

### Step 8: Solve for \( I \)
Thus,

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

### Final Result
The value of the integral is:

\[ \int_{0}^{1} \frac{\log(1+x)}{1+x^2} \, dx = \frac{\pi}{8} \log(2). \]