Let \[ I=\int \frac{e^{x}}{e^{4 x}+e^{2 x}+1} d x, \quad J=\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x \] Then, for an arbitrary constant $C$, the value of $J-I$ equals

(C) $\frac{1}{2} \log \left(\frac{e^{2 x}-e^{x}+1}{e^{2 x}+e^{x}+1}\right)+C$

(A) $\frac{1}{2} \log \left(\frac{e^{4 x}-e^{2 x}+1}{e^{4 x}+e^{2 x}+1}\right)+C$

(B) $\frac{1}{2} \log \left(\frac{e^{2 x}+e^{x}+1}{e^{2 x}-e^{x}+1}\right)+C$

(D) $\frac{1}{2} \log \left(\frac{e^{4 x}+e^{2 x}+1}{e^{4 x}-e^{2 x}+1}\right)+C$

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Indefinite Integrals

Let

I = \int \frac{e ^{x}}{e ^{4 x} + e ^{2 x} + 1} d x, J = \int \frac{e ^{- x}}{e ^{- 4 x} + e ^{- 2 x} + 1} d x

Then, for an arbitrary constant

C

, the value of

J - I

equals

(A) $\frac{1}{2} lo g (\frac{e ^{4 x} - e ^{2 x} + 1}{e ^{4 x} + e ^{2 x} + 1}) + C$
(B) $\frac{1}{2} lo g (\frac{e ^{2 x} + e ^{x} + 1}{e ^{2 x} - e ^{x} + 1}) + C$
(C) $\frac{1}{2} lo g (\frac{e ^{2 x} - e ^{x} + 1}{e ^{2 x} + e ^{x} + 1}) + C$ ✓ Correct
(D) $\frac{1}{2} lo g (\frac{e ^{4 x} + e ^{2 x} + 1}{e ^{4 x} - e ^{2 x} + 1}) + C$

Answer: (C)

We have

J - I = \int \frac{e ^{x} ( e ^{2 x} - 1 )}{e ^{4 x} + e ^{2 x} + 1} d x

. Let

z = e^{x}

, so

d z = e^{x} d x

. Then

J - I = \int \frac{z ^{2} - 1}{z ^{4} + z ^{2} + 1} d z = \int \frac{1 - z ^{- 2}}{z ^{2} + z ^{- 2} + 1} d z

. Let

w = z + z^{- 1}

, so

d w = (1 - z^{- 2}) d z

and

w^{2} = z^{2} + z^{- 2} + 2

. The integral becomes

\int \frac{d w}{w ^{2} - 1} = \frac{1}{2} ln \frac{w - 1}{w + 1} + C = \frac{1}{2} ln (\frac{e ^{x} + e ^{- x} - 1}{e ^{x} + e ^{- x} + 1}) + C = \frac{1}{2} ln (\frac{e ^{2 x} - e ^{x} + 1}{e ^{2 x} + e ^{x} + 1}) + C

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option