`int_1^3 (dx)/(x^2(1+x))`

Answer

To solve the integral \( I = \int_1^3 \frac{dx}{x^2(1+x)} \), we will follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral as an indefinite integral:
\[ I = \int \frac{dx}{x^2(1+x)} \]

### Step 2: Use Partial Fraction Decomposition
Next, we will perform partial fraction decomposition on the integrand:
\[ \frac{1}{x^2(1+x)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{1+x} \]
Multiplying through by the denominator \( x^2(1+x) \) gives:
\[ 1 = A x(1+x) + B(1+x) + C x^2 \]

### Step 3: Solve for Coefficients
Expanding the right-hand side:
\[ 1 = Ax + Ax^2 + B + Bx + Cx^2 \]
Combining like terms:
\[ 1 = (A + C)x^2 + (A + B)x + B \]
Setting coefficients equal gives us the system:
1. \( A + C = 0 \)
2. \( A + B = 0 \)
3. \( B = 1 \)

From \( B = 1 \), we find \( A + 1 = 0 \) which gives \( A = -1 \). Then from \( A + C = 0 \), we have \( -1 + C = 0 \) which gives \( C = 1 \).

Thus, we have:
\[ A = -1, \quad B = 1, \quad C = 1 \]

### Step 4: Rewrite the Integral
Now we can rewrite the integral:
\[ \int \left( \frac{-1}{x} + \frac{1}{x^2} + \frac{1}{1+x} \right) dx \]

### Step 5: Integrate Each Term
Now we integrate term by term:
\[ \int \left( -\frac{1}{x} + \frac{1}{x^2} + \frac{1}{1+x} \right) dx = -\ln|x| - \frac{1}{x} + \ln|1+x| + C \]

### Step 6: Evaluate the Definite Integral
Now we evaluate the definite integral from 1 to 3:
\[ I = \left[ -\ln|x| - \frac{1}{x} + \ln|1+x| \right]_1^3 \]

Calculating at the upper limit \( x = 3 \):
\[ -\ln(3) - \frac{1}{3} + \ln(4) \]

Calculating at the lower limit \( x = 1 \):
\[ -\ln(1) - 1 + \ln(2) = -1 + \ln(2) \]

### Step 7: Combine the Results
Now we combine the results:
\[ I = \left( -\ln(3) - \frac{1}{3} + \ln(4) \right) - \left( -1 + \ln(2) \right) \]
This simplifies to:
\[ I = -\ln(3) + \ln(4) + 1 - \ln(2) - \frac{1}{3} \]

Using properties of logarithms:
\[ \ln(4) - \ln(3) - \ln(2) = \ln\left(\frac{4}{3 \cdot 2}\right) = \ln\left(\frac{4}{6}\right) = \ln\left(\frac{2}{3}\right) \]

Thus:
\[ I = \ln\left(\frac{2}{3}\right) + 1 - \frac{1}{3} \]
\[ I = \ln\left(\frac{2}{3}\right) + \frac{2}{3} \]

### Final Result
The final result of the definite integral is:
\[ I = \frac{2}{3} \]