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

Answer

To solve the integral

\[ \int \frac{e^x (x^3 - x + 2)}{(1 + x^2)^2} \, dx, \]

we can follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting the integrand in a more manageable form. We can express \(x^3 - x + 2\) in a way that will allow us to use integration techniques effectively.

\[ x^3 - x + 2 = x^3 + x - 2x + 2 = x^3 + (x - 2x + 2) = x^3 + (1 + 1 - 2x) \]

This gives us:

\[ \int \frac{e^x (x^3 + (1 - 2x + 1))}{(1 + x^2)^2} \, dx \]

### Step 2: Split the integral
Next, we can split the integral into two parts:

\[ \int \frac{e^x x^3}{(1 + x^2)^2} \, dx + \int \frac{e^x (1 - 2x + 1)}{(1 + x^2)^2} \, dx \]

### Step 3: Simplify the second integral
Now, we simplify the second integral:

\[ \int \frac{e^x (2 - 2x)}{(1 + x^2)^2} \, dx = 2 \int \frac{e^x}{(1 + x^2)^2} \, dx - 2 \int \frac{e^x x}{(1 + x^2)^2} \, dx \]

### Step 4: Use substitution for the first integral
For the first integral, we can use integration by parts or a suitable substitution. Let's denote:

\[ f(x) = \frac{1 + x}{1 + x^2} \]

Then, we differentiate \(f(x)\):

\[ f'(x) = \frac{(1)(1 + x^2) - (2x)(1 + x)}{(1 + x^2)^2} = \frac{1 + x^2 - 2x - 2x^2}{(1 + x^2)^2} = \frac{1 - x^2 - 2x}{(1 + x^2)^2} \]

### Step 5: Identify the integral
Now, we can see that the integrand resembles \(e^x f'(x)\). Thus, we can apply the formula for the integral of the product of an exponential function and a function and its derivative:

\[ \int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C \]

### Step 6: Combine results
Combining the results, we have:

\[ \int \frac{e^x (x^3 - x + 2)}{(1 + x^2)^2} \, dx = e^x \left( \frac{1 + x}{1 + x^2} \right) + C \]

### Final Answer
Thus, the final result of the integral is:

\[ \int \frac{e^x (x^3 - x + 2)}{(1 + x^2)^2} \, dx = e^x \left( \frac{1 + x}{1 + x^2} \right) + C \]