`I= int ( 3x^(2) + 6x +5) "arc" tan x dx`.

Answer

To solve the integral \( I = \int (3x^2 + 6x + 5) \tan^{-1}(x) \, dx \), we will use 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:
- \( u = \tan^{-1}(x) \) (which is the inverse tangent function)
- \( dv = (3x^2 + 6x + 5) \, dx \)

### Step 2: Differentiate \( u \) and Integrate \( dv \)

Now we need to find \( du \) and \( v \):

1. Differentiate \( u \):
\[ du = \frac{1}{1+x^2} \, dx \]

2. Integrate \( dv \):
\[ v = \int (3x^2 + 6x + 5) \, dx = x^3 + 3x^2 + 5x \]

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

Substituting \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula:

\[ I = uv - \int v \, du \]
\[ I = \tan^{-1}(x) (x^3 + 3x^2 + 5x) - \int (x^3 + 3x^2 + 5x) \left(\frac{1}{1+x^2}\right) \, dx \]

### Step 4: Simplify the Integral

Now we need to simplify the integral:
\[ \int (x^3 + 3x^2 + 5x) \left(\frac{1}{1+x^2}\right) \, dx \]

This can be rewritten as:
\[ \int \left(\frac{x^3}{1+x^2} + \frac{3x^2}{1+x^2} + \frac{5x}{1+x^2}\right) \, dx \]

### Step 5: Divide and Simplify Each Term

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

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

3. For \( \frac{5x}{1+x^2} \):
\[ \frac{5x}{1+x^2} \text{ remains as is.} \]

### Step 6: Combine All Terms

Now, substituting these back, we have:
\[ \int \left( x - \frac{x}{1+x^2} + 3 - \frac{3}{1+x^2} + \frac{5x}{1+x^2} \right) \, dx \]
Combine the fractions:
\[ = \int \left( x + 3 + \frac{(5-1-3)x}{1+x^2} \right) \, dx = \int \left( x + 3 + \frac{x}{1+x^2} \right) \, dx \]

### Step 7: Integrate Each Term

Now, we can integrate each term:
1. \( \int x \, dx = \frac{x^2}{2} \)
2. \( \int 3 \, dx = 3x \)
3. \( \int \frac{x}{1+x^2} \, dx = \frac{1}{2} \ln(1+x^2) \)

### Step 8: Combine Everything

Putting it all together:
\[ I = \tan^{-1}(x)(x^3 + 3x^2 + 5x) - \left( \frac{x^2}{2} + 3x + \frac{1}{2} \ln(1+x^2) \right) + C \]

### Final Answer

Thus, the final answer is:
\[ I = \tan^{-1}(x)(x^3 + 3x^2 + 5x) - \left( \frac{x^2}{2} + 3x + \frac{1}{2} \ln(1+x^2) \right) + C \]