`int_(1)^(2) (1)/(x(1+x))dx`

Answer

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

### Step 1: Simplify the integrand
We start with the integrand \( \frac{1}{x(1+x)} \). We can use partial fraction decomposition to rewrite this expression:

\[ \frac{1}{x(1+x)} = \frac{A}{x} + \frac{B}{1+x} \]

Multiplying through by the denominator \( x(1+x) \) gives:

\[ 1 = A(1+x) + Bx \]

### Step 2: Solve for A and B
Expanding the right side:

\[ 1 = A + Ax + Bx \]

This can be rearranged to:

\[ 1 = A + (A + B)x \]

Now, we can set up a system of equations by equating coefficients:

1. \( A = 1 \)
2. \( A + B = 0 \)

From the first equation, we have \( A = 1 \). Substituting \( A \) into the second equation:

\[ 1 + B = 0 \implies B = -1 \]

Thus, we can rewrite the integrand:

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

### Step 3: Set up the integral
Now we can rewrite the integral:

\[ \int_{1}^{2} \frac{1}{x(1+x)} \, dx = \int_{1}^{2} \left( \frac{1}{x} - \frac{1}{1+x} \right) \, dx \]

### Step 4: Break the integral into two parts
We can now break this into two separate integrals:

\[ \int_{1}^{2} \frac{1}{x} \, dx - \int_{1}^{2} \frac{1}{1+x} \, dx \]

### Step 5: Compute the first integral
The integral of \( \frac{1}{x} \) is:

\[ \int \frac{1}{x} \, dx = \ln |x| + C \]

Evaluating from 1 to 2:

\[ \left[ \ln x \right]_{1}^{2} = \ln 2 - \ln 1 = \ln 2 \]

### Step 6: Compute the second integral
The integral of \( \frac{1}{1+x} \) is:

\[ \int \frac{1}{1+x} \, dx = \ln |1+x| + C \]

Evaluating from 1 to 2:

\[ \left[ \ln(1+x) \right]_{1}^{2} = \ln(3) - \ln(2) \]

### Step 7: Combine the results
Now we can combine the results of the two integrals:

\[ \int_{1}^{2} \frac{1}{x(1+x)} \, dx = \ln 2 - (\ln 3 - \ln 2) = \ln 2 - \ln 3 + \ln 2 = 2\ln 2 - \ln 3 \]

### Step 8: Use logarithmic properties
Using the properties of logarithms, we can simplify:

\[ 2\ln 2 - \ln 3 = \ln(2^2) - \ln 3 = \ln \left( \frac{4}{3} \right) \]

### Final Answer
Thus, the value of the definite integral is:

\[ \int_{1}^{2} \frac{1}{x(1+x)} \, dx = \ln \left( \frac{4}{3} \right) \]

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