`int(2x^(3))/((x^(2)+1)^(2))dx`

Answer

To solve the integral \( \int \frac{2x^3}{(x^2 + 1)^2} \, dx \), we will use the method of partial fractions. Here’s a step-by-step solution:

### Step 1: Set up the partial fraction decomposition
We can express the integrand as:
\[ \frac{2x^3}{(x^2 + 1)^2} = \frac{ax + b}{x^2 + 1} + \frac{cx + d}{(x^2 + 1)^2} \]

### Step 2: Multiply through by the denominator
Multiplying both sides by \( (x^2 + 1)^2 \) gives:
\[ 2x^3 = (ax + b)(x^2 + 1) + (cx + d) \]

### Step 3: Expand the right-hand side
Expanding the right-hand side:
\[ 2x^3 = ax^3 + bx^2 + ax + b + cx + d \]
This simplifies to:
\[ 2x^3 = ax^3 + (b + c)x^2 + (a + d)x + b \]

### Step 4: Compare coefficients
Now we compare coefficients from both sides:
- Coefficient of \( x^3 \): \( a = 2 \)
- Coefficient of \( x^2 \): \( b + c = 0 \)
- Coefficient of \( x \): \( a + d = 0 \)
- Constant term: \( b = 0 \)

From \( b = 0 \), we find:
- \( c = 0 \) (since \( b + c = 0 \))
- \( d = -2 \) (since \( a + d = 0 \))

Thus, we have:
\[ a = 2, \quad b = 0, \quad c = 0, \quad d = -2 \]

### Step 5: Rewrite the integral
Substituting back into the integral:
\[ \int \frac{2x^3}{(x^2 + 1)^2} \, dx = \int \left( \frac{2x}{x^2 + 1} - \frac{2}{(x^2 + 1)^2} \right) \, dx \]

### Step 6: Integrate each term
Now we can integrate each term separately:
1. For \( \int \frac{2x}{x^2 + 1} \, dx \):
- Let \( t = x^2 + 1 \), then \( dt = 2x \, dx \).
- Thus, \( \int \frac{2x}{x^2 + 1} \, dx = \int \frac{1}{t} \, dt = \ln |t| + C = \ln(x^2 + 1) + C \).

2. For \( \int \frac{2}{(x^2 + 1)^2} \, dx \):
- This integral can be solved using the formula \( \int \frac{1}{(x^2 + a^2)^2} \, dx = \frac{x}{2a^2(x^2 + a^2)} + C \).
- Here, \( a = 1 \), so:
\[ \int \frac{2}{(x^2 + 1)^2} \, dx = -\frac{1}{x^2 + 1} + C \]

### Step 7: Combine the results
Putting it all together:
\[ \int \frac{2x^3}{(x^2 + 1)^2} \, dx = \ln(x^2 + 1) - \frac{1}{x^2 + 1} + C \]

### Final Answer
Thus, the final result is:
\[ \int \frac{2x^3}{(x^2 + 1)^2} \, dx = \ln(x^2 + 1) - \frac{1}{x^2 + 1} + C \]