`int (tan ^-1 x^2) x dx`

Answer

To solve the integral \( I = \int \tan^{-1}(x^2) \cdot x \, dx \), we will use integration by parts and a substitution method. Here’s a step-by-step solution:

### Step 1: Set up the integral
Let \( I = \int \tan^{-1}(x^2) \cdot x \, dx \).

### Step 2: Use substitution
Let \( t = x^2 \). Then, differentiate to find \( dt \):
\[ dt = 2x \, dx \quad \Rightarrow \quad x \, dx = \frac{dt}{2}. \]
Now, rewrite the integral in terms of \( t \):
\[ I = \int \tan^{-1}(t) \cdot \frac{dt}{2} = \frac{1}{2} \int \tan^{-1}(t) \, dt. \]

### Step 3: Apply integration by parts
For the integral \( \int \tan^{-1}(t) \, dt \), we will use integration by parts. Let:
- \( u = \tan^{-1}(t) \) (thus \( du = \frac{1}{1+t^2} \, dt \))
- \( dv = dt \) (thus \( v = t \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int \tan^{-1}(t) \, dt = t \tan^{-1}(t) - \int t \cdot \frac{1}{1+t^2} \, dt. \]

### Step 4: Simplify the remaining integral
The remaining integral is:
\[ \int \frac{t}{1+t^2} \, dt. \]
This can be simplified by using the substitution \( w = 1 + t^2 \), giving \( dw = 2t \, dt \) or \( dt = \frac{dw}{2t} \):
\[ \int \frac{t}{1+t^2} \, dt = \frac{1}{2} \int \frac{1}{w} \, dw = \frac{1}{2} \ln |w| + C = \frac{1}{2} \ln(1+t^2) + C. \]

### Step 5: Combine results
Now, substituting back into the integration by parts result:
\[ \int \tan^{-1}(t) \, dt = t \tan^{-1}(t) - \frac{1}{2} \ln(1+t^2) + C. \]
Thus, we have:
\[ I = \frac{1}{2} \left( t \tan^{-1}(t) - \frac{1}{2} \ln(1+t^2) \right) + C. \]

### Step 6: Substitute back for \( t \)
Recall that \( t = x^2 \):
\[ I = \frac{1}{2} \left( x^2 \tan^{-1}(x^2) - \frac{1}{2} \ln(1+x^4) \right) + C. \]

### Final Answer
The final result of the integral is:
\[ I = \frac{1}{2} x^2 \tan^{-1}(x^2) - \frac{1}{4} \ln(1+x^4) + C. \]