The value of `int_(1)^(3)(dx)/(x(1+x^(2)))` is

Options

A. `1/2log(9/5)`
B. `1/2 log(5/9)`
C. `1/2 log(4/9)`
D. `1/2log(9/4)`

Answer

To solve the integral \( I = \int_{1}^{3} \frac{dx}{x(1+x^2)} \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_{1}^{3} \frac{dx}{x(1+x^2)} \]

### Step 2: Simplify the Integrand
We can simplify the integrand by using partial fraction decomposition. We want to express:
\[ \frac{1}{x(1+x^2)} = \frac{A}{x} + \frac{Bx + C}{1+x^2} \]
Multiplying through by the denominator \(x(1+x^2)\) gives:
\[ 1 = A(1+x^2) + (Bx + C)x \]
Expanding this, we have:
\[ 1 = A + Ax^2 + Bx^2 + Cx \]
Combining like terms, we get:
\[ 1 = (A + B)x^2 + Cx + A \]
From this, we can equate coefficients:
1. \(A = 1\)
2. \(B + A = 0 \Rightarrow B = -1\)
3. \(C = 0\)

Thus, we can rewrite our integrand as:
\[ \frac{1}{x(1+x^2)} = \frac{1}{x} - \frac{1}{1+x^2} \]

### Step 3: Rewrite the Integral
Now we can rewrite the integral \(I\):
\[ I = \int_{1}^{3} \left( \frac{1}{x} - \frac{1}{1+x^2} \right) dx \]

### Step 4: Split the Integral
We can split the integral into two parts:
\[ I = \int_{1}^{3} \frac{1}{x} \, dx - \int_{1}^{3} \frac{1}{1+x^2} \, dx \]

### Step 5: Evaluate the First Integral
The first integral is:
\[ \int \frac{1}{x} \, dx = \ln|x| \]
Thus,
\[ \int_{1}^{3} \frac{1}{x} \, dx = \ln(3) - \ln(1) = \ln(3) \]

### Step 6: Evaluate the Second Integral
The second integral is:
\[ \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) \]
Thus,
\[ \int_{1}^{3} \frac{1}{1+x^2} \, dx = \tan^{-1}(3) - \tan^{-1}(1) = \tan^{-1}(3) - \frac{\pi}{4} \]

### Step 7: Combine the Results
Now we can combine the results from both integrals:
\[ I = \ln(3) - \left( \tan^{-1}(3) - \frac{\pi}{4} \right) \]
This simplifies to:
\[ I = \ln(3) + \frac{\pi}{4} - \tan^{-1}(3) \]

### Final Answer
Thus, the value of the integral is:
\[ I = \ln(3) + \frac{\pi}{4} - \tan^{-1}(3) \]