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. $\frac{1}{2}log\left(\frac{e^{2x}+e^x+1}{e^{2x}-e^x+1}\right)+c$
• C. $\frac{1}{2}log\left(\frac{e^{2x}-e^{2x}+1}{e^{2x}+e^x+1}\right)+c$
• D. $\frac{1}{2}log\left(\frac{e^{4x}-e2x+1}{e^{4x}-e^{2x}+1}\right)+c$

Answer 1

The correct option is C

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

Given $I=\int \frac{e^x}{e^{4x}+e^{2x}+1}dxJ=\int \frac{e^{-x}}{e^{-4x}+e^{-2x}+1}dx=\int \frac{e^{3x}}{e^{4x}+e^{2x}+1}\therefore J-I=\int \frac{e^x(e^{2x}-1)}{e^{4x}+e^{2x}+1}dx\text{Let}{e}^x=t\Rightarrow e^{x}dx=dt\therefore J-I=\int \frac{t^2-1}{t^4+t^2+1}dt=\int \frac{1-\frac{1}{t^2}}{t^2+1+\frac{1}{t^2}}dt\text{Let}t+\frac{1}{t}=u\Rightarrow (1-\frac{1}{t^2})dt=du\therefore J-I=\int \frac{du}{u^2-1}=\frac{1}{2}log\left|\frac{u-1}{u+1}\right|+c=\frac{1}{2}log\left|\frac{\frac{t^2+1}{t}-1}{\frac{t^2+1}{t}+1}\right|+c=\frac{1}{2}log\left|\frac{e^{2x-e^x+1}}{e^{2x}+e^x+1}\right|+c$