Evaluate :
`int x^(3)tan^(-1)x dx`

Answer

To evaluate the integral \( \int x^3 \tan^{-1}(x) \, dx \), we will use the method of integration by parts. The formula for integration by parts is given by:

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

### Step 1: Choose \( u \) and \( dv \)
Let:
- \( u = \tan^{-1}(x) \) (first function)
- \( dv = x^3 \, dx \) (second 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 \( dv \):
\[ v = \int x^3 \, dx = \frac{x^4}{4} \]

### Step 3: Apply the integration by parts formula
Now we apply the integration by parts formula:
\[ \int x^3 \tan^{-1}(x) \, dx = uv - \int v \, du \]
Substituting the values we found:
\[ = \tan^{-1}(x) \cdot \frac{x^4}{4} - \int \frac{x^4}{4} \cdot \frac{1}{1+x^2} \, dx \]
This simplifies to:
\[ = \frac{x^4}{4} \tan^{-1}(x) - \frac{1}{4} \int \frac{x^4}{1+x^2} \, dx \]

### Step 4: Simplify the integral
Now we simplify the integral \( \int \frac{x^4}{1+x^2} \, dx \):
\[ \frac{x^4}{1+x^2} = \frac{x^4 - 1 + 1}{1+x^2} = \frac{(x^2 - 1)(x^2 + 1)}{1+x^2} + 1 \]
This can be rewritten as:
\[ = x^2 - 1 + \frac{1}{1+x^2} \]
Thus, we have:
\[ \int \frac{x^4}{1+x^2} \, dx = \int (x^2 - 1) \, dx + \int \frac{1}{1+x^2} \, dx \]

### Step 5: Evaluate the integrals
Now we evaluate the integrals:
1. \( \int (x^2 - 1) \, dx = \frac{x^3}{3} - x \)
2. \( \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) \)

Putting it all together:
\[ \int \frac{x^4}{1+x^2} \, dx = \left( \frac{x^3}{3} - x + \tan^{-1}(x) \right) \]

### Step 6: Substitute back into the equation
Substituting this back into our equation:
\[ \int x^3 \tan^{-1}(x) \, dx = \frac{x^4}{4} \tan^{-1}(x) - \frac{1}{4} \left( \frac{x^3}{3} - x + \tan^{-1}(x) \right) \]

### Step 7: Simplify the expression
This simplifies to:
\[ = \frac{x^4}{4} \tan^{-1}(x) - \frac{x^3}{12} + \frac{x}{4} - \frac{1}{4} \tan^{-1}(x) + C \]

### Final Result
Combining the terms gives us the final result:
\[ \int x^3 \tan^{-1}(x) \, dx = \left( \frac{x^4}{4} - \frac{1}{4} \right) \tan^{-1}(x) - \frac{x^3}{12} + \frac{x}{4} + C \]