`int\ cot^(-1)x\ dx`

Answer

To solve the integral \( \int \cot^{-1}(x) \, dx \), we will use integration by parts. Here’s the step-by-step solution:

### Step 1: Identify \( u \) and \( dv \)
We will use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = \cot^{-1}(x) \) (inverse function)
- \( dv = dx \) (algebraic function)

### Step 2: Differentiate \( u \) and Integrate \( dv \)
Now we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = -\frac{1}{1 + x^2} \, dx \]
- Integrate \( dv \):
\[ v = x \]

### Step 3: Apply the Integration by Parts Formula
Substituting \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula:
\[ \int \cot^{-1}(x) \, dx = x \cot^{-1}(x) - \int x \left(-\frac{1}{1 + x^2}\right) \, dx \]
This simplifies to:
\[ \int \cot^{-1}(x) \, dx = x \cot^{-1}(x) + \int \frac{x}{1 + x^2} \, dx \]

### Step 4: Solve the Remaining Integral
Now we need to solve the integral \( \int \frac{x}{1 + x^2} \, dx \). We can use substitution:
Let \( t = 1 + x^2 \), then \( dt = 2x \, dx \) or \( dx = \frac{dt}{2x} \).

Thus, we can rewrite the integral:
\[ \int \frac{x}{1 + x^2} \, dx = \frac{1}{2} \int \frac{1}{t} \, dt \]
This integral evaluates to:
\[ \frac{1}{2} \ln |t| + C = \frac{1}{2} \ln(1 + x^2) + C \]

### Step 5: Combine All Parts
Now, substitute back into our expression:
\[ \int \cot^{-1}(x) \, dx = x \cot^{-1}(x) + \frac{1}{2} \ln(1 + x^2) + C \]

### Final Answer
Thus, the final result is:
\[ \int \cot^{-1}(x) \, dx = x \cot^{-1}(x) + \frac{1}{2} \ln(1 + x^2) + C \]

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