The value of the integral `int_(1/2)^2 tan^-1 x/x dx` is equal to

Options

A. `pi log_e2`
B. `1/2log_e2`
C. `pi/2log_e2`
D. `pi/4log_e2`

Answer

To solve the integral \( I = \int_{\frac{1}{2}}^{2} \frac{\tan^{-1} x}{x} \, dx \), we can use a substitution technique to simplify the integral. Here’s a step-by-step solution:

### Step 1: Define the Integral
Let
\[ I = \int_{\frac{1}{2}}^{2} \frac{\tan^{-1} x}{x} \, dx \]

### Step 2: Substitute \( x \) with \( \frac{1}{x} \)
Now, we will substitute \( x \) with \( \frac{1}{x} \). When \( x = \frac{1}{2} \), \( \frac{1}{x} = 2 \) and when \( x = 2 \), \( \frac{1}{x} = \frac{1}{2} \). The differential \( dx \) changes as follows:
\[ dx = -\frac{1}{x^2} \, dx \]

Thus, the integral becomes:
\[ I = \int_{2}^{\frac{1}{2}} \frac{\tan^{-1} \left(\frac{1}{x}\right)}{\frac{1}{x}} \left(-\frac{1}{x^2}\right) \, dx \]
This simplifies to:
\[ I = \int_{\frac{1}{2}}^{2} \frac{\tan^{-1} \left(\frac{1}{x}\right)}{x} \, dx \]

### Step 3: Use the Identity for \( \tan^{-1} \)
We know that:
\[ \tan^{-1} \left(\frac{1}{x}\right) + \tan^{-1} x = \frac{\pi}{2} \]
Thus, we can rewrite the integral:
\[ I = \int_{\frac{1}{2}}^{2} \frac{\frac{\pi}{2} - \tan^{-1} x}{x} \, dx \]

### Step 4: Split the Integral
Now we can split the integral:
\[ I = \int_{\frac{1}{2}}^{2} \frac{\frac{\pi}{2}}{x} \, dx - \int_{\frac{1}{2}}^{2} \frac{\tan^{-1} x}{x} \, dx \]
This gives us:
\[ I = \frac{\pi}{2} \int_{\frac{1}{2}}^{2} \frac{1}{x} \, dx - I \]

### Step 5: Solve for \( I \)
Now, we can solve for \( I \):
\[ 2I = \frac{\pi}{2} \int_{\frac{1}{2}}^{2} \frac{1}{x} \, dx \]
Calculating the integral:
\[ \int_{\frac{1}{2}}^{2} \frac{1}{x} \, dx = \ln x \bigg|_{\frac{1}{2}}^{2} = \ln 2 - \ln \frac{1}{2} = \ln 2 + \ln 2 = 2 \ln 2 \]
So,
\[ 2I = \frac{\pi}{2} \cdot 2 \ln 2 = \pi \ln 2 \]
Thus,
\[ I = \frac{\pi}{2} \ln 2 \]

### Final Answer
The value of the integral is:
\[ \boxed{\frac{\pi}{2} \ln 2} \]