`int x^(3) "arc" tan x dx`.

Answer

To solve 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:

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

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

### Step 3: Apply the Integration by Parts Formula
Substituting into the integration by parts formula:
\[ \int x^3 \tan^{-1}(x) \, dx = \left( \tan^{-1}(x) \cdot \frac{x^4}{4} \right) - \int \left( \frac{x^4}{4} \cdot \frac{1}{1 + x^2} \right) \, dx \]
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 \( \int \frac{x^4}{1 + x^2} \, dx \):
\[ \frac{x^4}{1 + x^2} = \frac{x^4 + x^2 - x^2}{1 + x^2} = x^2 - \frac{x^2}{1 + x^2} \]
Thus,
\[ \int \frac{x^4}{1 + x^2} \, dx = \int x^2 \, dx - \int \frac{x^2}{1 + x^2} \, dx \]

### Step 5: Solve Each Integral
1. The first integral:
\[ \int x^2 \, dx = \frac{x^3}{3} \]

2. The second integral can be solved by using the substitution \( w = 1 + x^2 \):
\[ \int \frac{x^2}{1 + x^2} \, dx = \int \left( 1 - \frac{1}{1 + x^2} \right) \, dx = x - \tan^{-1}(x) \]

### Step 6: Combine Results
Putting it all together:
\[ \int \frac{x^4}{1 + x^2} \, dx = \frac{x^3}{3} - \left( x - \tan^{-1}(x) \right) = \frac{x^3}{3} - x + \tan^{-1}(x) \]

### Step 7: Substitute Back
Substituting back into our integration by parts result:
\[ \int x^3 \tan^{-1}(x) \, dx = \frac{x^4}{4} \tan^{-1}(x) - \frac{1}{4} \left( \frac{x^3}{3} - x + \tan^{-1}(x) \right) \]

### Step 8: Final Simplification
\[ = \frac{x^4}{4} \tan^{-1}(x) - \frac{x^3}{12} + \frac{x}{4} - \frac{1}{4} \tan^{-1}(x) + C \]
Thus, the final answer is:
\[ \int x^3 \tan^{-1}(x) \, dx = \left( \frac{x^4}{4} - \frac{1}{4} \right) \tan^{-1}(x) - \frac{x^3}{12} + \frac{x}{4} + C \]