Question

Let I = $\int_{}^{}\frac{e^{x}}{e^{4x} + e^{2x} + 1}$ dx, J = $\int_{}^{}\frac{e^{–x}}{e^{–4x} + e^{–2x} + 1}$ dx. Then, for an arbitrary constant C, the value of J – I equals –

• A. $\frac{1}{2}$log$\left( \frac{e^{4x}–e^{2x} + 1}{e^{4x} + e^{2x} + 1} \right)$ + C
• B. $\frac{1}{2}$log$\left( \frac{e^{2x} + e^{x} + 1}{e^{2x}–e^{2x} + 1} \right)$ + C
• C. $\frac{1}{2}$log$\left( \frac{e^{2x}–e^{x} + 1}{e^{2x} + e^{x} + 1} \right)$ + C
• D. $\frac{1}{2}$log$\left( \frac{e^{4x} + e^{2x} + 1}{e^{4x}–e^{2x} + 1} \right)$ + C

Answer 1

Answer: (C)

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

Answer 2

J – I = $\int_{}^{}\frac{e^{3x}–e^{x}}{1 + e^{2x} + e^{4x}}$

Let e^x = t

̃ e^xdx = dt

J-I = $\int_{}^{}\frac{t^{2}–1}{1 + t^{2} + t^{4}}$dt = $\int_{}^{}{\frac{t^{2}\left( 1–\frac{1}{t^{2}} \right)}{t^{2}\left( t^{2} + 1 + \frac{1}{t^{2}} \right)}dt}$

= dt

t + = u ̃ $\left( 1–\frac{1}{t^{2}} \right)$dt = du

=

= $\frac{1}{2}$log $\left| \frac{u–1}{u + 1} \right|$ + c

= $\frac{1}{2}$ log $\left| \frac{e^{x} + \frac{1}{e^{x}}–1}{e^{x} + \frac{1}{e^{x}} + 1} \right|$ + c