`intxtan^(-1)x dx =`

Options

A. `1/2(x^2+1)tan^(-1)x-x`
B. `1/2(x^2+1)tan^(-1)x-1/2x`
C. `(x^2+1) tan^(-1)x - x`
D. None of these

Answer

To solve the integral \( \int x \tan^{-1}(x) \, dx \), we will use the method of integration by parts. The formula for integration by parts is:

\[ \int u \, dv = uv - \int v \, du \]

### Step 1: Choose \( u \) and \( dv \)

Let:
- \( u = \tan^{-1}(x) \) (which we will differentiate)
- \( dv = x \, dx \) (which we will integrate)

### 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 \( dv \):
\[ v = \frac{x^2}{2} \]

### Step 3: Apply the Integration by Parts Formula

Now we can apply the integration by parts formula:
\[ \int x \tan^{-1}(x) \, dx = uv - \int v \, du \]
Substituting the values we found:
\[ = \tan^{-1}(x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{1+x^2} \, dx \]

### Step 4: Simplify the Integral

Now simplify the integral:
\[ = \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2} \int \frac{x^2}{1+x^2} \, dx \]
We can simplify \( \frac{x^2}{1+x^2} \):
\[ \frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2} \]
Thus, we can rewrite the integral:
\[ = \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2} \left( \int 1 \, dx - \int \frac{1}{1+x^2} \, dx \right) \]

### Step 5: Integrate

Now, we can integrate:
\[ \int 1 \, dx = x \]
\[ \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) \]
Putting it all together:
\[ = \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2} \left( x - \tan^{-1}(x) \right) \]
\[ = \frac{x^2}{2} \tan^{-1}(x) - \frac{x}{2} + \frac{1}{2} \tan^{-1}(x) \]

### Step 6: Combine Like Terms

Now, combine the terms:
\[ = \left( \frac{x^2}{2} + \frac{1}{2} \right) \tan^{-1}(x) - \frac{x}{2} \]

### Final Answer

Thus, the final answer is:
\[ \int x \tan^{-1}(x) \, dx = \frac{x^2 + 1}{2} \tan^{-1}(x) - \frac{x}{2} + C \]