Evaluate: `int((xtan^(-1)x)/((1+x^2)^(3setminus2))\) dx`

Answer

To evaluate the integral

\[ I = \int \frac{x \tan^{-1}(x)}{(1+x^2)^{\frac{3}{2}}} \, dx, \]

we will follow these steps:

### Step 1: Rewrite the Denominator
We can express the denominator as:

\[ (1+x^2)^{\frac{3}{2}} = (1+x^2)^{1} \cdot (1+x^2)^{\frac{1}{2}}. \]

Thus, we can rewrite the integral as:

\[ I = \int \frac{x \tan^{-1}(x)}{(1+x^2) \cdot (1+x^2)^{\frac{1}{2}}} \, dx. \]

### Step 2: Substitution
Let \( y = \tan^{-1}(x) \). Then, we have:

\[ x = \tan(y) \quad \text{and} \quad dx = \frac{1}{1+x^2} \, dy. \]

Substituting these into the integral gives:

\[ I = \int \frac{\tan(y) \cdot y}{(1+\tan^2(y))^{\frac{3}{2}}} \cdot \frac{1}{1+\tan^2(y)} \, dy. \]

### Step 3: Simplifying the Integral
Using the identity \( 1 + \tan^2(y) = \sec^2(y) \), we can rewrite the integral:

\[ I = \int \frac{y \tan(y)}{\sec^3(y)} \cdot \frac{1}{\sec^2(y)} \, dy = \int y \sin(y) \, dy. \]

### Step 4: Integration by Parts
We will use integration by parts, where we let:

- \( u = y \) and \( dv = \sin(y) \, dy \).
- Then, \( du = dy \) and \( v = -\cos(y) \).

Applying integration by parts:

\[ I = -y \cos(y) + \int \cos(y) \, dy. \]

### Step 5: Evaluate the Remaining Integral
The integral of \( \cos(y) \) is:

\[ \int \cos(y) \, dy = \sin(y). \]

Thus, we have:

\[ I = -y \cos(y) + \sin(y) + C. \]

### Step 6: Substitute Back
Now we substitute back \( y = \tan^{-1}(x) \):

\[ I = -\tan^{-1}(x) \cos(\tan^{-1}(x)) + \sin(\tan^{-1}(x)) + C. \]

### Step 7: Simplify Using Right Triangle
Using a right triangle where the opposite side is \( x \) and the adjacent side is \( 1 \):

- The hypotenuse is \( \sqrt{1+x^2} \).
- Therefore, \( \cos(\tan^{-1}(x)) = \frac{1}{\sqrt{1+x^2}} \) and \( \sin(\tan^{-1}(x)) = \frac{x}{\sqrt{1+x^2}} \).

Substituting these values gives:

\[ I = -\tan^{-1}(x) \cdot \frac{1}{\sqrt{1+x^2}} + \frac{x}{\sqrt{1+x^2}} + C. \]

### Final Answer
Thus, the evaluated integral is:

\[ I = \frac{x - \tan^{-1}(x)}{\sqrt{1+x^2}} + C. \]