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

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

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

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

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

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

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

We have,

$\begin{array}{l} J - I = \int \frac{e^{3 x}}{e^{\frac{4 x}{2 x} + e^{2 x} + 1}} - \frac{e^{x}}{e^{4 x} + e^{2 x} + 1} d x \\ \Rightarrow J - I = \int \frac{e^{2 x} - 1}{e^{4 x} + e^{2 x} + 1} d (e^{x}) \end{array}$

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

$\begin{aligned} \Rightarrow J - I = \int \frac{1}{{(t + \frac{1}{t})}^{2} - 1} d (t - \frac{1}{t}) \\ \Rightarrow J - I = \frac{1}{2} \log (\frac{t + \frac{1}{t} - 1}{t + \frac{t}{t} + 1}) + C \end{aligned}$

$\Rightarrow J - I = \frac{1}{2} \log (\frac{t^{2} - t + 1}{t^{2} + t + 1}) + C = \frac{1}{2} \log (\frac{e^{2 x} - e^{x} + 1}{e^{2 x} + e^{x} + 1}) + C$