If I = ∫ frace^xe^4x + e^2x + 1 dx, J = ∫ frace^-xe^-4x +… — JEE Main 2008 Mathematics PYQ

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.
$\frac{1}{2} \log \left| \frac{e^{2x} + e^{x} + 1}{e^{2x} - e^{x} + 1} \right| + c$
C.
$\frac{1}{2} \log \left| \frac{e^{2x} - e^{x} + 1}{e^{2x} + e^{x} + 1} \right| + c$
D.
$\frac{1}{2} \log \left| \frac{e^{4x} + e^{2x} + 1}{e^{4x} - e^{2x} + 1} \right| + c$

Show answer and solution

Correct option: C

Solution

Rewrite J by multiplying numerator and denominator by e^(4x):

J = ∫ e^(3x)/(e^(4x) + e^(2x) + 1) dx

So J - I = ∫ (e^(3x) - e^x)/(e^(4x) + e^(2x) + 1) dx = ∫ e^x(e^(2x) - 1)/(e^(4x) + e^(2x) + 1) dx

Substitute t = e^x, dt = e^x dx:

J - I = ∫ (t² - 1)/(t⁴ + t² + 1) dt

Divide numerator and denominator by t²:

= ∫ (1 - 1/t²)/(t² + 1 + 1/t²) dt = ∫ (1 - 1/t²)/((t + 1/t)² - 1) dt

Let u = t + 1/t, then du = (1 - 1/t²) dt:

= ∫ du/(u² - 1) = (1/2) log|(u-1)/(u+1)| + c = (1/2) log|(t + 1/t - 1)/(t + 1/t + 1)| + c

Substituting back t = e^x and multiplying inside by e^x:

= (1/2) log|(e^(2x) - e^x + 1)/(e^(2x) + e^x + 1)| + c

[AIEEE 2008]

JEE Main 2008 Mathematics question on Indefinite Integration