Question

Let $I=\int \frac{e^x}{e^{4x}+e^{2x}+1}ex,J=\int \frac{e^{-x}}{e^{-4x}+e^{-2x}+1}dx.$ Then, for an arbitrary constant c, the value of J - I equals

• A. $\frac{1}{2}\ell n\left(\frac{e^{4x}-e^{2x}+1}{e^{4x}+e^{2x}+1}\right)+c$
• B. $\frac{1}{2}\ell n\left(\frac{e^{2x}+e^x+1}{e^{2x}-e^x+1}\right)+c$
• C. $\frac{1}{2}\ell n\left(\frac{e^{2x}-e^x+1}{e^{2x}+e^x+1}\right)+c$
• D. $\frac{1}{2}\ell n\left(\frac{e^{4x}+e^{2x}+1}{e^{4x}-e^{2x}+1}\right)+c$

Answer 1

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

Since $I=\int \frac{e^x}{e^{4x}+e^{2x}+1}dx$ and $J=\int \frac{e^{3x}}{1+e^{2x}+e^{4x}}dx$

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

Substitute $e^x=u\Rightarrow e^{x}dx=du$

$\therefore J-I=\int \frac{(u^2-1)}{1+u^2+u^4}du=\int \frac{(1-\frac{1}{u^2})}{1+\frac{1}{u^2}+u^2}du$

$=\int \frac{(1-\frac{1}{u^2})}{1+\frac{1}{u^2}+u^2}du$

Substitute $u+\frac{1}{u}=t\Rightarrow (1-\frac{1}{u^2})dx=dt$

$J-I=\int \frac{dt}{t^2-1}$

$=\frac{1}{2}\log \left|\frac{t-1}{t+1}\right|+c=\frac{1}{2}\log \left|\frac{u^2-u+1}{u^2+u+1}\right|+c$

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