`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 \( (x^2 - 1) + 2 \) to facilitate the integration. Thus, we have:

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

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

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

### Step 3: Simplify the first integral
The first integral can be simplified using the identity \( x^2 - 1 = (x + 1)(x - 1) \):

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

### Step 4: Use integration by parts
Now we can apply the integration by parts formula. Let \( u = \frac{x - 1}{x + 1} \) and \( dv = e^x \, dx \). Then we need to find \( du \):

\[ u = \frac{x - 1}{x + 1} \implies du = \frac{(x + 1)(1) - (x - 1)(1)}{(x + 1)^2} \, dx = \frac{2}{(x + 1)^2} \, dx \]

Now we compute \( v \):

\[ v = e^x \]

### Step 5: Apply integration by parts
Using integration by parts:

\[ \int u \, dv = uv - \int v \, du \]

So we have:

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

### Step 6: Combine the integrals
Now we can combine the integrals:

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

The two \( \int e^x \frac{2}{(x + 1)^2} \, dx \) terms cancel out, leaving us with:

\[ = e^x \frac{x - 1}{x + 1} + C \]

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

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