Question

If I$=\int \limits \frac {e^x}{e^{4x+e^{2x+1}}}dx, J=\int \limits \frac {e^{-x}}{e^{-4x}+e^{-2x}+1} dx$ 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 $=\int \limits \frac {e^x}{e^{4x}+e^{2x+1}}dx$
and J$=\int \limits \frac {e^{3x}}{1+e^{2x+}e^{4x}}dx$
$\therefore \, \, \, \, \, \, J-I = \int \limits \frac {(e^{3x}-e^x)}{1+e^{2x}+e^{4x}}dx$
Put $e^x=u \Rightarrow e^xdx=du$
$\therefore \, \, \, \, J-I= \int \limits \frac {(u^2-1)}{1+u^2+u^4}du=\int \limits \frac {\bigg (1-\frac {1}{u^2}\bigg )}{1+ \frac {1}{u^2}+u^2}du$
$ = \int \limits \frac {\bigg (1-\frac {1}{u^2}\bigg )}{\bigg (u+ \frac {1}{u}\bigg )^2-1}du$
put $u+ \frac {1}{u}=t \Rightarrow \bigg (1- \frac {1}{u^2}\bigg )du=dt$
$ =\int \limits \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$