Key Concepts and Formulas

• Integration by Parts: $\int u \, dv = uv - \int v \, du$
• A useful form for integration: $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$
• Basic differentiation rules: Power rule, constant rule, etc.

Step-by-Step Solution

Step 1: Rewrite the integrand.

We want to manipulate the expression inside the integral to match a form that is easily integrable, specifically aiming for the form $e^x [f(x) + f'(x)]$.

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

Step 2: Separate the fraction.

We split the fraction into two parts to simplify the expression.

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

Step 3: Factor the first term.

We factor the numerator of the first term using the difference of squares.

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

Step 4: Simplify the first term.

We cancel the common factor $(x+1)$.

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

Step 5: Combine the integrals.

We combine the two integrals into a single integral.

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

Step 6: Identify $f(x)$ and $f'(x)$.

We want to express the integrand in the form $e^x [f(x) + f'(x)]$. Let's assume $f(x) = \frac{x-1}{x+1}$. Then we need to find $f'(x)$.

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

Now we can see that the integrand is indeed in the form $e^x [f(x) + f'(x)]$.

Step 7: Evaluate the integral.

Using the formula $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$, we have:

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

Step 8: Identify $f(x)$.

From the given equation, we have:

$\int \frac{(x^2 + 1)e^x}{(x+1)^2} dx = f(x)e^x + C$

Comparing this with our result, we get:

$f(x) = \frac{x-1}{x+1}$

Step 9: Find the first derivative, $f'(x)$.

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

Step 10: Find the second derivative, $f''(x)$.

$f''(x) = 2(-2)(x+1)^{-3}(1) = -4(x+1)^{-3} = \frac{-4}{(x+1)^3}$

Step 11: Find the third derivative, $f'''(x)$.

$f'''(x) = -4(-3)(x+1)^{-4}(1) = 12(x+1)^{-4} = \frac{12}{(x+1)^4}$

Step 12: Evaluate $f'''(1)$.

$f'''(1) = \frac{12}{(1+1)^4} = \frac{12}{2^4} = \frac{12}{16} = \frac{3}{4}$

Common Mistakes & Tips

• Algebraic Manipulation: Be careful with algebraic manipulations, especially when dealing with fractions. Double-check your work to avoid errors.
• Recognizing the Form: The key to solving this problem is recognizing the form $\int e^x [f(x) + f'(x)] \, dx$. Practice identifying such patterns in integrals.
• Differentiation: Remember the chain rule when differentiating composite functions.

Summary

We started by rewriting the integrand to get it into a form that we could easily integrate. We identified $f(x)$ and $f'(x)$, integrated, and then found the first, second, and third derivatives of $f(x)$. Finally, we evaluated the third derivative at $x=1$.

The final answer is $\boxed{3/4}$, which corresponds to option (B).