Definite Integration
Indefinite Integration
Grade Class 12
Question:
Let I = ∫ \frac{e^x}{e^{4x} + e^{2x} + 1} dx, J = ∫ \frac{e^{-x}}{e^{-4x} + e^{-2x} + 1} dx. Then, for an arbitrary constant c, the value of J - I equals
(A) \frac{1}{2} \ln \left| \frac{e^{4x} - e^{2x} + 1}{e^{4x} + e^{2x} + 1} \right| + c
(B) \frac{1}{2} \ln \left| \frac{e^{2x} + e^x + 1}{e^{2x} - e^x + 1} \right| + c
(C) \frac{1}{2} \ln \left| \frac{e^{2x} - e^x + 1}{e^{2x} + e^x + 1} \right| + c
(D) \frac{1}{2} \ln \left| \frac{e^{4x} + e^x + 1}{e^{4x} - e^{2x} + 1} \right| + c
Step-by-Step Solution
Key Concept: Simplify J by multiplying numerator and denominator by e^4x to express it in terms of e^x, then compute J - I.
J = \int \frac{e^{-x}}{e^{-4x} + e^{-2x} + 1} dx = \int \frac{e^{-x} \cdot e^{4x}}{e^{-4x} \cdot e^{4x} + e^{-2x} \cdot e^{4x} + 1 \cdot e^{4x}} dx = \int \frac{e^{3x}}{1 + e^{2x} + e^{4x}} dx. Let u = e^x, du = e^x dx. Then I = \int \frac{1}{u^4 + u^2 + 1} du and J = \int \frac{u^2}{u^4 + u^2 + 1} du. J - I = \int \frac{u^2 - 1}{u^4 + u^2 + 1} du = \int \frac{1 - 1/u^2}{u^2 + 1 + 1/u^2} du = \int \frac{1 - 1/u^2}{(u + 1/u)^2 - 1} du. Let t = u + 1/u, dt = (1 - 1/u^2) du. Then J - I = \int \frac{dt}{t^2 - 1} = \frac{1}{2} \ln |\frac{t-1}{t+1}| + c = \frac{1}{2} \ln |\frac{u + 1/u - 1}{u + 1/u + 1}| + c = \frac{1}{2} \ln |\frac{u^2 - u + 1}{u^2 + u + 1}| + c = \frac{1}{2} \ln |\frac{e^{2x} - e^x + 1}{e^{2x} + e^x + 1}| + c.
Correct Answer: C