Step 1 (Formula / Governing Law & Given Data): [1/4 Mark]

We are given the definite integral:

$$I = \int_{0}^{1} \frac{dx}{e^x + e^{-x}}$$

Using the algebraic property $e^{-x} = \frac{1}{e^x}$, we rewrite the integrand as:

$$I = \int_{0}^{1} \frac{e^x}{e^{2x} + 1} dx$$

We utilize the standard integration formula: $\int \frac{du}{u^2 + 1} = \tan^{-1}(u) + C$.

Step 2 (Step-by-Step Mathematical Working): [1/2 Mark]

Let $u = e^x$. Then, the derivative is $du = e^x dx$.

Change the limits of integration:

• When $x = 0$, $u = e^0 = 1$.
• When $x = 1$, $u = e^1 = e$.

Substituting these into the integral:

$$I = \int_{1}^{e} \frac{du}{u^2 + 1}$$

Evaluating the integral:

$$I = \left[ \tan^{-1}(u) \right]_{1}^{e}$$ $$I = \tan^{-1}(e) - \tan^{-1}(1)$$

Since $\tan^{-1}(1) = \frac{\pi}{4}$, we have:

$$I = \tan^{-1}(e) - \frac{\pi}{4}$$

Step 3 (Final Concluded Answer): [1/4 Mark]

Comparing the result $I = \tan^{-1}(e) - \frac{\pi}{4}$ with the given expression $\tan^{-1}(e) + k$:

$$k = -\frac{\pi}{4}$$

Therefore, the correct option is (D).

💡 EL10 Concept Insight: By multiplying the top and bottom by $e^x$, we turn a messy fraction into a classic "derivative over function squared plus one" pattern, which is just the secret code for the inverse tangent function!