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 for $J-I$ 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. $\frac{1}{2}\log \left(\frac{e^{4x}+e^{2x}+1}{e^{4x}-e^{2x}+1}\right)+C$

Answer 1

The correct option is C $\frac{1}{2}\log \left(\frac{e^{2x}-e^x+1}{e^{2x}+e^x+1}\right)+C$

Given $I=\int \frac{e^x}{e^{4x}+e^{2x}+1}dx$,
$J=\int \frac{e^{-x}}{e^{-4x}+e^{-2x}+1}dx=\int \frac{e^{3x}}{e^{4x}+e^{2x}+1}$
$J-I=\int \frac{e^x(e^{2x}-1)}{e^{4x}+e^{2x}+1}dx$

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