Question

Evaluate \( \int_{ \log 4}^{ \log 5} \frac{e^{2x} + e^x}{e^{2x} - 5e^x +6} dx \):

• A. \( \log \left( \frac{64}{9} \right) \)
• B. \( \log \left( \frac{256}{81} \right) \)
• C. \( \log \left( \frac{32}{3} \right) \)
• D. \( \log \left( \frac{128}{27} \right) \)

Hint:

Substituting \( t = e^x \) in integrals with exponentials simplifies the problem into algebraic fractions.

Answer 1

The Correct Option is D

Step 1: Substituting \( t = e^x \). Let \( t = e^x \), then \( dt = e^x dx = t dx \). Thus, changing the limits: \[ x = \log 4 \Rightarrow t = 4, \quad x = \log 5 \Rightarrow t = 5. \] Rewriting the integral in terms of \( t \): \[ I = \int_{4}^{5} \frac{t^2 + t}{t^2 - 5t + 6} dt. \] Step 2: Partial Fraction Decomposition. Factoring the denominator: \[ t^2 - 5t + 6 = (t-2)(t-3). \] Using partial fractions and solving the integral step-by-step gives: \[ I = \log \left( \frac{128}{27} \right). \]