Question

If 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. $\frac{1}{2}log|\frac{e^{4x}-e^{2x}+1}{e^{4x}+e^{2x}+1}|+c$
• B. $\frac{1}{2}log|\frac{e^{2x}+e^x+1}{e^{2x}-e^x+1}|+c$
• C. $\frac{1}{2}log|\frac{e^{2x}-e^x+1}{e^{2x}+e^x+1}|+c$
• D. $\frac{1}{2}log|\frac{e^{4x}+e^{2x}+1}{e^{4x}-e^{2x}+1}|+c$

Answer 1

Answer: (C)

Since, I $=\int \frac{e^x}{e^{4x}+e^{2x+1}}dx$
and J$=\int \frac{e^{3x}}{1+e^{2x+}{e}^{4x}}dx$
$\therefore J-I=\int \frac{(e^{3x}-e^x)}{1+e^{2x}+e^{4x}}dx$
Put $e^x=u\Rightarrow e^{x}dx=du$
$\therefore J-I=\int \frac{(u^2-1)}{1+u^2+u^4}du=\int \frac{(1-\frac{1}{u^2})}{1+\frac{1}{u^2}+u^2}du$
$=\int \frac{(1-\frac{1}{u^2})}{(u+\frac{1}{u}{)}^2-1}du$
put $u+\frac{1}{u}=t\Rightarrow (1-\frac{1}{u^2})du=dt$
$=\int \frac{dt}{t^2-1}=\frac{1}{2}log|\frac{t-1}{t+1}|+c$
$=\frac{1}{2}log|\frac{u^2-u+1}{u^2+u+1}|+c=\frac{1}{2}log|\frac{e^{2x}-e^x+1}{e^{2x}+e^x+1}|+c$