`int_(0)^(oo) (xdx)/((1+x) (1+x^(2)))=`

Options

A. `(pi)/(4)`
B. `(pi)/(2)`
C. `pi`
D. none

Answer

To solve the integral

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

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

### Step 1: Substitution
Let \( x = \tan(\theta) \). Then, the differential \( dx \) is given by:

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

### Step 2: Change of Limits
Now, we change the limits of integration:
- When \( x = 0 \), \( \theta = \tan^{-1}(0) = 0 \).
- When \( x = \infty \), \( \theta = \tan^{-1}(\infty) = \frac{\pi}{2} \).

### Step 3: Substitute into the Integral
Now we substitute \( x \) and \( dx \) into the integral:

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

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

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

### Step 4: Simplify the Integral
We can rewrite \( \tan(\theta) \) as \( \frac{\sin(\theta)}{\cos(\theta)} \):

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\frac{\sin(\theta)}{\cos(\theta)}}{1 + \frac{\sin(\theta)}{\cos(\theta)}} \, d\theta = \int_{0}^{\frac{\pi}{2}} \frac{\sin(\theta)}{\sin(\theta) + \cos(\theta)} \, d\theta. \]

### Step 5: Use Symmetry
Now, we can use the property of integrals. We know that:

\[ \int_{0}^{\frac{\pi}{2}} \frac{\sin(\theta)}{\sin(\theta) + \cos(\theta)} \, d\theta + \int_{0}^{\frac{\pi}{2}} \frac{\cos(\theta)}{\sin(\theta) + \cos(\theta)} \, d\theta = \int_{0}^{\frac{\pi}{2}} 1 \, d\theta = \frac{\pi}{2}. \]

Let \( J = \int_{0}^{\frac{\pi}{2}} \frac{\cos(\theta)}{\sin(\theta) + \cos(\theta)} \, d\theta \). Then we have:

\[ I + J = \frac{\pi}{2}. \]

### Step 6: Evaluate \( J \)
By substituting \( \theta = \frac{\pi}{2} - u \):

\[ J = \int_{0}^{\frac{\pi}{2}} \frac{\sin(u)}{\sin(u) + \cos(u)} \, du = I. \]

Thus, we have:

\[ I + I = \frac{\pi}{2} \implies 2I = \frac{\pi}{2} \implies I = \frac{\pi}{4}. \]

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

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