Question

If \(\int_{0}^{1} \frac{dx}{e^x + e^{-x}} = \tan^{-1}e + k\), then the constant value of \(k\) is:

• A. \(e\)
• B. \frac{\pi}{4}
• C. \(0\)
• D. \(-\frac{\pi}{4}\)

Hint:

Never forget to convert the integration limits when doing variable substitutions in definite integrals! Changing limits early keeps calculations clean and prevents errors from back-substitution at the end.

Answer 1

The Correct Option is D

Concept: To solve a definite integral involving exponential terms of opposite signs, such as \(e^x\) and \(e^{-x}\), we rewrite the negative exponent as a fraction: \[ e^{-x} = \frac{1}{e^x} \] This rational transition enables us to apply a substitution approach by setting \(u = e^x\), which simplifies the calculation into a standard rational algebraic integral matching the form \(\int \frac{du}{1+u^2} = \tan^{-1}u\).

Step 1: Simplify the algebraic integrand formula
Let the definite integral on the left-hand side be denoted by \(I\): \[ I = \int_{0}^{1} \frac{dx}{e^x + \frac{1}{e^x}} \] Find a common denominator for the terms in the denominator: \[ e^x + \frac{1}{e^x} = \frac{(e^x)^2 + 1}{e^x} = \frac{e^{2x} + 1}{e^x} \] Substituting this back inverted into the integral yields: \[ I = \int_{0}^{1} \frac{e^x dx}{e^{2x} + 1} = \int_{0}^{1} \frac{e^x dx}{(e^x)^2 + 1} \]

Step 2: Use substitution method and transform boundaries
Let \(u = e^x\). Differentiating both sides with respect to \(x\): \[ du = e^x dx \] Now, we must adjust our definite integration lower and upper bounds from \(x\) coordinates to our new tracking variable \(u\):
• When lower bound \(x = 0\): \(u = e^0 = 1\)
• When upper bound \(x = 1\): \(u = e^1 = e\)

Step 3: Execute integration using the new parameters
Reconstructing the definite integral completely using the substituted variable \(u\): \[ I = \int_{1}^{e} \frac{du}{u^2 + 1} \] The standard antiderivative for this expression is well-known: \[ I = \Big[ \tan^{-1} u \Big]_{1}^{e} \] Applying the fundamental theorem of calculus by evaluating at the upper and lower limits: \[ I = \tan^{-1}(e) - \tan^{-1}(1) \] We know from standard trigonometric exact values that \(\tan\left(\frac{\pi}{4}\right) = 1\), which means \(\tan^{-1}(1) = \frac{\pi}{4}\). \[ I = \tan^{-1}(e) - \frac{\pi}{4} \]

Step 4: Equate and compute the value of \(k\)
The problem statement gives the following relation condition: \[ I = \tan^{-1}e + k \] Equating our newly calculated analytical result to this given expression form: \[ \tan^{-1}(e) - \frac{\pi}{4} = \tan^{-1}(e) + k \] Subtracting the common term \(\tan^{-1}(e)\) from both sides leaves: \[ k = -\frac{\pi}{4} \] This corresponds exactly to choice option (D).