Answer :
To solve the integral [tex]\( J = \int_1^2 \frac{\ln (1+t)}{t^2} \, dt \)[/tex], we will go through a detailed, step-by-step solution.
1. Identify the integral: The integral we want to solve is
[tex]\[ J = \int_1^2 \frac{\ln(1+t)}{t^2} \, dt. \][/tex]
2. Choose an appropriate method: In this case, we will use integration by parts. To apply integration by parts, we use the formula:
[tex]\[ \int u \, dv = uv - \int v \, du. \][/tex]
Here, we need to choose [tex]\( u \)[/tex] and [tex]\( dv \)[/tex] carefully. Let's set:
[tex]\[ u = \ln(1+t) \quad \text{and} \quad dv = \frac{1}{t^2} \, dt. \][/tex]
We then need to find [tex]\( du \)[/tex] and [tex]\( v \)[/tex]:
- Differentiate [tex]\( u = \ln(1+t) \)[/tex]:
[tex]\[ du = \frac{d}{dt} [\ln(1+t)] \, dt = \frac{1}{1+t} \, dt. \][/tex]
- Integrate [tex]\( dv = \frac{1}{t^2} \, dt \)[/tex]:
[tex]\[ v = \int \frac{1}{t^2} \, dt = -\frac{1}{t}. \][/tex]
3. Apply the integration by parts formula:
[tex]\[ J = \int_1^2 \ln(1+t) \cdot \frac{1}{t^2} \, dt = \left. \ln(1+t) \cdot \left( -\frac{1}{t} \right) \right|_1^2 - \int_1^2 \left( -\frac{1}{t} \right) \cdot \frac{1}{1+t} \, dt. \][/tex]
Simplify the boundary term:
[tex]\[ \left. -\frac{\ln(1+t)}{t} \right|_1^2 = -\left(\frac{\ln(1+2)}{2} - \frac{\ln(1+1)}{1} \right) = -\left(\frac{\ln 3}{2} - \ln 2 \right). \][/tex]
4. Simplify the remaining integral:
Now we will evaluate the remaining integral:
[tex]\[ -\int_1^2 \frac{1}{t(1+t)} \, dt = -\int_1^2 \frac{1}{t(1+t)} \, dt. \][/tex]
We can use partial fractions for this part:
[tex]\[ \frac{1}{t(1+t)} = \frac{A}{t} + \frac{B}{1+t}. \][/tex]
Solving for [tex]\( A \)[/tex] and [tex]\( B \)[/tex] gives:
[tex]\[ 1 = A(1+t) + Bt \implies A = 1 \quad \text{and} \quad B = -1. \][/tex]
Thus,
[tex]\[ \frac{1}{t(1+t)} = \frac{1}{t} - \frac{1}{1+t}. \][/tex]
So,
[tex]\[ -\int_1^2 \left( \frac{1}{t} - \frac{1}{1+t} \right) \, dt = -\left( \int_1^2 \frac{1}{t} \, dt - \int_1^2 \frac{1}{1+t} \, dt \right). \][/tex]
Evaluating these integrals:
[tex]\[ \int_1^2 \frac{1}{t} \, dt = \ln t \bigg|_1^2 = \ln 2 - \ln 1 = \ln 2. \][/tex]
[tex]\[ \int_1^2 \frac{1}{1+t} \, dt = \ln (1+t) \bigg|_1^2 = \ln 3 - \ln 2. \][/tex]
5. Combine the results:
Putting it all together, we get:
[tex]\[ J = -\left( \frac{\ln 3}{2} - \ln 2 \right) - \left( \ln 2 - (\ln 3 - \ln 2) \right). \][/tex]
Simplify the expressions:
[tex]\[ J = -\left( \frac{\ln 3}{2} - \ln 2 \right) - \ln 2 + \ln 3 - \ln 2. \][/tex]
[tex]\[ J = -\frac{\ln 3}{2} + \ln 2 - \ln 2 + \ln 3 - \ln 2. \][/tex]
[tex]\[ J = -\frac{\ln 3}{2} + \ln 3 - 2\ln 2. \][/tex]
The computed numerical value for this integral is approximately:
[tex]\[ J \approx 0.431523108677671. \][/tex]
1. Identify the integral: The integral we want to solve is
[tex]\[ J = \int_1^2 \frac{\ln(1+t)}{t^2} \, dt. \][/tex]
2. Choose an appropriate method: In this case, we will use integration by parts. To apply integration by parts, we use the formula:
[tex]\[ \int u \, dv = uv - \int v \, du. \][/tex]
Here, we need to choose [tex]\( u \)[/tex] and [tex]\( dv \)[/tex] carefully. Let's set:
[tex]\[ u = \ln(1+t) \quad \text{and} \quad dv = \frac{1}{t^2} \, dt. \][/tex]
We then need to find [tex]\( du \)[/tex] and [tex]\( v \)[/tex]:
- Differentiate [tex]\( u = \ln(1+t) \)[/tex]:
[tex]\[ du = \frac{d}{dt} [\ln(1+t)] \, dt = \frac{1}{1+t} \, dt. \][/tex]
- Integrate [tex]\( dv = \frac{1}{t^2} \, dt \)[/tex]:
[tex]\[ v = \int \frac{1}{t^2} \, dt = -\frac{1}{t}. \][/tex]
3. Apply the integration by parts formula:
[tex]\[ J = \int_1^2 \ln(1+t) \cdot \frac{1}{t^2} \, dt = \left. \ln(1+t) \cdot \left( -\frac{1}{t} \right) \right|_1^2 - \int_1^2 \left( -\frac{1}{t} \right) \cdot \frac{1}{1+t} \, dt. \][/tex]
Simplify the boundary term:
[tex]\[ \left. -\frac{\ln(1+t)}{t} \right|_1^2 = -\left(\frac{\ln(1+2)}{2} - \frac{\ln(1+1)}{1} \right) = -\left(\frac{\ln 3}{2} - \ln 2 \right). \][/tex]
4. Simplify the remaining integral:
Now we will evaluate the remaining integral:
[tex]\[ -\int_1^2 \frac{1}{t(1+t)} \, dt = -\int_1^2 \frac{1}{t(1+t)} \, dt. \][/tex]
We can use partial fractions for this part:
[tex]\[ \frac{1}{t(1+t)} = \frac{A}{t} + \frac{B}{1+t}. \][/tex]
Solving for [tex]\( A \)[/tex] and [tex]\( B \)[/tex] gives:
[tex]\[ 1 = A(1+t) + Bt \implies A = 1 \quad \text{and} \quad B = -1. \][/tex]
Thus,
[tex]\[ \frac{1}{t(1+t)} = \frac{1}{t} - \frac{1}{1+t}. \][/tex]
So,
[tex]\[ -\int_1^2 \left( \frac{1}{t} - \frac{1}{1+t} \right) \, dt = -\left( \int_1^2 \frac{1}{t} \, dt - \int_1^2 \frac{1}{1+t} \, dt \right). \][/tex]
Evaluating these integrals:
[tex]\[ \int_1^2 \frac{1}{t} \, dt = \ln t \bigg|_1^2 = \ln 2 - \ln 1 = \ln 2. \][/tex]
[tex]\[ \int_1^2 \frac{1}{1+t} \, dt = \ln (1+t) \bigg|_1^2 = \ln 3 - \ln 2. \][/tex]
5. Combine the results:
Putting it all together, we get:
[tex]\[ J = -\left( \frac{\ln 3}{2} - \ln 2 \right) - \left( \ln 2 - (\ln 3 - \ln 2) \right). \][/tex]
Simplify the expressions:
[tex]\[ J = -\left( \frac{\ln 3}{2} - \ln 2 \right) - \ln 2 + \ln 3 - \ln 2. \][/tex]
[tex]\[ J = -\frac{\ln 3}{2} + \ln 2 - \ln 2 + \ln 3 - \ln 2. \][/tex]
[tex]\[ J = -\frac{\ln 3}{2} + \ln 3 - 2\ln 2. \][/tex]
The computed numerical value for this integral is approximately:
[tex]\[ J \approx 0.431523108677671. \][/tex]
