1. Key Concepts and Formulas

• Substitution Method for Integration: Used to simplify integrals by changing the variable of integration. The differential element $dx$ must also be transformed into $dt$ using the chain rule.
• Standard Integral Forms: Knowledge of common integration formulas, specifically $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$.
• Properties of Quadratic Equations: A quadratic equation with roots $r_1$ and $r_2$ can be expressed as $x^2 - (r_1+r_2)x + r_1r_2 = 0$.
• Inverse Trigonometric Functions: Understanding values of inverse trigonometric functions like $\tan^{-1}(\sqrt{3})$ and $\tan^{-1}(1)$.

2. Step-by-Step Solution

We are given the definite integral: \int_\limits\alpha^{\log_e 4} \frac{\mathrm{d} x}{\sqrt{\mathrm{e}^x-1}}=\frac{\pi}{6}

Step 1: Perform a Substitution to Simplify the Integral To simplify the integrand $\frac{1}{\sqrt{e^x-1}}$, we use the substitution $t = \sqrt{e^x-1}$.

• Why this substitution? This substitution is effective for integrals involving $\sqrt{e^x - \text{constant}}$ as it directly removes the square root. Squaring both sides gives $t^2 = e^x - 1$. Rearranging, we get $e^x = t^2 + 1$. Differentiating both sides with respect to $x$ using the chain rule: $\frac{\mathrm{d}}{\mathrm{d}x}(e^x) = \frac{\mathrm{d}}{\mathrm{d}x}(t^2+1)$ $e^x = 2t \frac{\mathrm{d}t}{\mathrm{d}x}$ Solving for $dx$: $\mathrm{d}x = \frac{2t}{e^x} \mathrm{d}t$ Substituting $e^x = t^2+1$: $\mathrm{d}x = \frac{2t}{t^2+1} \mathrm{d}t$ Now, substitute $t$ and $dx$ into the integral: $\int \frac{1}{t} \left(\frac{2t}{t^2+1}\right) \mathrm{d}t = \int \frac{2}{t^2+1} \mathrm{d}t$

Step 2: Evaluate the Indefinite Integral The integral $\int \frac{2}{t^2+1} \mathrm{d}t$ is a standard form. $2 \int \frac{1}{t^2+1} \mathrm{d}t = 2 \tan^{-1}(t) + C$

• Why this formula? This uses the standard integral $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$ with $a=1$. Substitute back $t = \sqrt{e^x-1}$: $\int \frac{\mathrm{d} x}{\sqrt{\mathrm{e}^x-1}} = 2 \tan^{-1}(\sqrt{e^x-1})$

Step 3: Apply the Limits of Integration Now we evaluate the definite integral using the given limits $\alpha$ and $\log_e 4$. \left[2 \tan^{-1}(\sqrt{e^x-1})\right]_\limits\alpha^{\log_e 4} = \frac{\pi}{6} Evaluate at the upper limit $x = \log_e 4$: $2 \tan^{-1}(\sqrt{e^{\log_e 4}-1}) = 2 \tan^{-1}(\sqrt{4-1}) = 2 \tan^{-1}(\sqrt{3})$ Since $\tan(\frac{\pi}{3}) = \sqrt{3}$, we have $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$. So, the upper limit evaluation is $2 \times \frac{\pi}{3} = \frac{2\pi}{3}$. Evaluate at the lower limit $x = \alpha$: $2 \tan^{-1}(\sqrt{e^\alpha-1})$ Now, subtract the lower limit value from the upper limit value: $\frac{2\pi}{3} - 2 \tan^{-1}(\sqrt{e^\alpha-1}) = \frac{\pi}{6}$

Step 4: Solve for $\alpha$ We need to isolate the term involving $\alpha$. $2 \tan^{-1}(\sqrt{e^\alpha-1}) = \frac{2\pi}{3} - \frac{\pi}{6}$ Find a common denominator for the right side: $2 \tan^{-1}(\sqrt{e^\alpha-1}) = \frac{4\pi}{6} - \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$ Divide by 2: $\tan^{-1}(\sqrt{e^\alpha-1}) = \frac{\pi}{4}$ Take the tangent of both sides: $\sqrt{e^\alpha-1} = \tan\left(\frac{\pi}{4}\right)$ Since $\tan(\frac{\pi}{4}) = 1$: $\sqrt{e^\alpha-1} = 1$ Square both sides: $e^\alpha-1 = 1$ $e^\alpha = 2$ Now, we can find $e^{-\alpha}$: $e^{-\alpha} = \frac{1}{e^\alpha} = \frac{1}{2}$

Step 5: Form the Quadratic Equation We are looking for a quadratic equation whose roots are $e^\alpha$ and $e^{-\alpha}$. Let $r_1 = e^\alpha = 2$ and $r_2 = e^{-\alpha} = \frac{1}{2}$. The sum of the roots is $r_1 + r_2 = 2 + \frac{1}{2} = \frac{5}{2}$. The product of the roots is $r_1 r_2 = 2 \times \frac{1}{2} = 1$. A quadratic equation with roots $r_1$ and $r_2$ is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$. Substituting the sum and product: $x^2 - \frac{5}{2}x + 1 = 0$ Multiply the entire equation by 2 to eliminate the fraction: $2x^2 - 5x + 2 = 0$

Common Mistakes & Tips

• Chain Rule Application: Ensure the chain rule is correctly applied when differentiating implicitly or changing variables. For example, $\frac{d}{dx}(t^2) = 2t \frac{dt}{dx}$.
• Trigonometric Values: Memorize common values for trigonometric and inverse trigonometric functions, such as $\tan^{-1}(\sqrt{3})=\frac{\pi}{3}$ and $\tan^{-1}(1)=\frac{\pi}{4}$.
• Algebraic Simplification: Pay close attention to algebraic manipulations, especially when dealing with fractions and exponents, to avoid calculation errors.

3. Summary

The problem involves evaluating a definite integral using a trigonometric substitution. After simplifying the integral and applying the limits, we solved for $e^\alpha$. The values of $e^\alpha$ and $e^{-\alpha}$ were then used to construct the required quadratic equation based on the relationship between roots and coefficients. The roots were found to be $2$ and $\frac{1}{2}$, leading to the quadratic equation $2x^2 - 5x + 2 = 0$.

The final answer is \boxed{\text{2 x^2-5 x+2=0}}, which corresponds to option (A).