The value of the integration \(\int \frac{dx}{(1+e^{x})(1-e^{-x})}\) is

\(-\frac{1}{1+e^{2}}\)

The correct answer is Option 4: \(-\frac{1}{1+e^{2}}\)

Given integral:

$\displaystyle I=\int \frac{dx}{(1+e^x)(1-e^{-x})}$

Rewrite $1-e^{-x}=\frac{e^x-1}{e^x}$:

$\displaystyle I=\int \frac{dx}{(1+e^x)\frac{e^x-1}{e^x}} =\int \frac{e^x}{e^{2x}-1}\,dx$

Put $t=e^x \Rightarrow dt=e^x dx$:

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

Factor:

$t^2-1=(t-1)(t+1)$

Using partial fractions:

$\displaystyle \frac{1}{t^2-1}=\frac12\left(\frac{1}{t-1}-\frac{1}{t+1}\right)$

Integrate:

$\displaystyle I=\frac12\ln|t-1|-\frac12\ln|t+1|+C$

$\displaystyle I=\frac12\ln\left|\frac{t-1}{t+1}\right|+C$

Substitute $t=e^x$:

$\displaystyle I=\frac12\ln\left|\frac{e^x-1}{e^x+1}\right|+C$

This is the correct antiderivative.

The expression $-\frac{1}{1+x^2}$ is not the solution of this integral.

Final answer: $\displaystyle \frac12\ln\left|\frac{e^x-1}{e^x+1}\right|+C$