Step 1: Understanding the Concept:
This problem represents a classic standard form in indefinite integration involving the exponential function \( e^{x} \).
The structure \( \int e^{x} [f(x) + f'(x)] dx \) is very common in competitive exams.
Step 2: Key Formula or Approach:
The standard integral identity is:
\[ \int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C \]
This can be proven using integration by parts on the first term: \( \int e^{x} f(x) dx \).
Let \( u = f(x) \) and \( dv = e^{x} dx \). Then \( du = f'(x) dx \) and \( v = e^{x} \).
By parts: \( \int e^{x} f(x) dx = f(x) e^{x} - \int e^{x} f'(x) dx \).
Adding the \( \int e^{x} f'(x) dx \) term from the original integral cancels the second part, leaving just \( e^{x} f(x) \).
Step 3: Detailed Explanation:
We are given the integral:
\[ \int e^{x} \left( \tan^{-1} x + \frac{1}{1+x^{2}} \right) dx \]
Let's define the parts:
Let \( f(x) = \tan^{-1} x \).
Now, calculate the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^{2}} \]
The given integrand is exactly in the form \( e^{x} [f(x) + f'(x)] \).
Substituting these into our standard formula:
\[ I = e^{x} f(x) + C \]
\[ I = e^{x} \tan^{-1} x + C \]
This derivation is straightforward once the pattern is recognized.
The other options are incorrect because they either add unnecessary constants (a), miss the exponential term (b), or use the wrong trigonometric inverse function (d).
Step 4: Final Answer:
The solution to the integral is \( e^{x} \tan^{-1} x + C \).