Let I=∫exe4x+e2x+1dx,J=∫e−xe−4x+e−2x+1dx, Then, for an arbitrary constant C, the value of J-I equals

# 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

1. A

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

2. B

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

3. C

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

4. D

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

Register to Get Free Mock Test and Study Material

+91

Verify OTP Code (required)

### Solution:

We have,

where $t={e}^{x}$

Register to Get Free Mock Test and Study Material

+91

Verify OTP Code (required)