Evaluate: `int (e^(x) (x^(2) + 1))/((x+1)^(2))dx`

Answer

To evaluate the integral

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

we can use the technique of integration by parts and the known formula for integrals involving \( e^x \).

### Step 1: Rewrite the Integral
We can rewrite the integrand by manipulating the expression:

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

This allows us to separate the integral into two parts:

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

### Step 2: Factor the First Integral
Notice that \( x^2 - 1 \) can be factored as \( (x - 1)(x + 1) \):

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

This simplifies to:

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

### Step 3: Apply Integration by Parts
For the first integral \( \int e^x \frac{x - 1}{x + 1} \, dx \), we can use integration by parts. Let:

- \( u = \frac{x - 1}{x + 1} \) and \( dv = e^x \, dx \).

Then, we need to find \( du \) and \( v \):

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

Now, applying integration by parts:

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

we get:

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

### Step 4: Combine the Integrals
Now substitute back into the expression for \( I \):

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

The integrals involving \( \int e^x \frac{2}{(x + 1)^2} \, dx \) will cancel out:

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

Thus, the integral simplifies to:

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

where \( C \) is the constant of integration.

### Final Answer
The evaluated integral is:

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