`int e^(x)((x^(2)+1))/((x+1)^(2))dx` is equal to

Options

A. `((x-1)/(x+1))e^(x)+c`
B. `e^(x)((x+1)/(x-1))+c`
C. `e^(x)(x+1)(x-1)+c`
D. none of these

Explanation

To solve the integral \( \int e^x \frac{x^2 + 1}{(x + 1)^2} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We start with the expression:
\[ \frac{x^2 + 1}{(x + 1)^2} \]
We can rewrite \( x^2 + 1 \) as:
\[ x^2 + 1 = (x^2 - 1) + 2 = (x - 1)(x + 1) + 2 \]
Thus, we can express the integral as:
\[ \int e^x \left( \frac{(x - 1)(x + 1)}{(x + 1)^2} + \frac{2}{(x + 1)^2} \right) \, dx \]

### Step 2: Simplify the expression
Now, we can simplify the integrand:
\[ \int e^x \left( \frac{x - 1}{x + 1} + \frac{2}{(x + 1)^2} \right) \, dx \]

### Step 3: Split the integral
We can split the integral into two parts:
\[ \int e^x \frac{x - 1}{x + 1} \, dx + \int e^x \frac{2}{(x + 1)^2} \, dx \]

### Step 4: Apply integration by parts
For the first integral \( \int e^x \frac{x - 1}{x + 1} \, dx \), we can use the identity:
\[ \int e^x f(x) \, dx = e^x f(x) + \int e^x f'(x) \, dx \]
Let \( f(x) = \frac{x - 1}{x + 1} \).

### Step 5: Differentiate \( f(x) \)
Using the quotient rule:
\[ f'(x) = \frac{(x + 1)(1) - (x - 1)(1)}{(x + 1)^2} = \frac{2}{(x + 1)^2} \]

### Step 6: Substitute back into the integral
Now, we can substitute back into our integral:
\[ \int e^x \frac{x - 1}{x + 1} \, dx = e^x \frac{x - 1}{x + 1} - \int e^x \frac{2}{(x + 1)^2} \, dx \]

### Step 7: Combine the integrals
Now, we can combine the two integrals:
\[ \int e^x \frac{x^2 + 1}{(x + 1)^2} \, dx = e^x \frac{x - 1}{x + 1} + C \]

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