A natural exponential function in terms of $x$ involved in a mathematical relation to form a rational expression. We have to evaluate the indefinite integration of the rational expression with respect to $x$ in this calculus problem.
$\displaystyle \int{\dfrac{1}{1+e^x} \,} dx$
To evaluate the integral of the exponential rational function, we have to simplify the expression firstly.
Take $z = 1+e^x$ and differentiate this equation with respect to $x$.
$\implies$ $\dfrac{d}{dx}{\,(z)} \,=\, \dfrac{d}{dx}{\,(1+e^x)}$
$\implies$ $\dfrac{d}{dx}{\,(z)} \,=\, \dfrac{d}{dx}{\,(1)}+\dfrac{d}{dx}{\,(e^x)}$
$\implies$ $\dfrac{dz}{dx} \,=\, 0+e^x$
$\implies$ $\dfrac{dz}{dx} \,=\, e^x$
$\implies$ $dz \,=\, e^x \times dx$
$\implies$ $dx \,=\, \dfrac{dz}{e^x}$
Now, convert the rational expression in terms of $z$ from $x$.
$\implies$ $dx \,=\, \dfrac{dz}{z-1}$
$\implies$ $\displaystyle \int{\dfrac{1}{1+e^x} \,} dx$ $\,=\,$ $\displaystyle \int{\dfrac{1}{z} \,} \Bigg(\dfrac{dz}{z-1}\Bigg)$
$= \,\,\,$ $\displaystyle \int{\dfrac{1}{z(z-1)} \,} dz$
The algebraic expression in rational form represents a rational expression that consists of non-repeated linear factors in the denominator. So, it can be decomposed into sum of the partial fractions.
$\implies$ $\displaystyle \int{\dfrac{1}{z(z-1)} \,} dz$ $\,=\,$ $\displaystyle \int{\Bigg(\dfrac{A}{z}+\dfrac{B}{z-1}\Bigg) \,} dz$
Put $z = 0$, then $A$ $\,=\,$ $\dfrac{1}{0-1}$ $\,=\,$ $\dfrac{1}{-1}$ $\,=\,$ $-1$
Put $z = 1$, then $B$ $\,=\,$ $\dfrac{1}{1}$ $\,=\,$ $1$
Now, substitute the values of $A$ and $B$ to complete the partial fraction decomposition.
$\implies$ $\displaystyle \int{\dfrac{1}{z(z-1)} \,} dz$ $\,=\,$ $\displaystyle \int{\Bigg(\dfrac{-1}{z}+\dfrac{1}{z-1}\Bigg) \,} dz$
$\implies$ $\displaystyle \int{\dfrac{1}{z(z-1)} \,} dz$ $\,=\,$ $\displaystyle \int{\Bigg(-\dfrac{1}{z}+\dfrac{1}{z-1}\Bigg) \,} dz$
$\implies$ $\displaystyle \int{\dfrac{1}{z(z-1)} \,} dz$ $\,=\,$ $\displaystyle \int{\Bigg(\dfrac{1}{z-1}-\dfrac{1}{z}\Bigg) \,} dz$
We can now evaluate the indefinite integral of the expression with respect to $x$.
$\implies$ $\displaystyle \int{\Bigg(\dfrac{1}{z-1}-\dfrac{1}{z}\Bigg) \,} dz$ $\,=\,$ $\displaystyle \int{\dfrac{1}{z-1}\,} dz$ $-$ $\displaystyle \int{\dfrac{1}{z}\,} dz$
The indefinite integration of the second integral term can be evaluated by the reciprocal rule of integration.
$=\,\,\,$ $\displaystyle \int{\dfrac{1}{z-1}\,} dz$ $-$ $\log_e{z}+c_2$
Now, take $y = z-1$ and differentiate the equation both sides with respect to $z$.
$\implies$ $\dfrac{d}{dz}{\, (y)}$ $\,=\,$ $\dfrac{d}{dz}{\, (z-1)}$
$\implies$ $\dfrac{dy}{dz}$ $\,=\,$ $\dfrac{d}{dz}{\, (z)}-\dfrac{d}{dz}{\,(1)}$
$\implies$ $\dfrac{dy}{dz}$ $\,=\,$ $1-0$
$\implies$ $\dfrac{dy}{dz}$ $\,=\,$ $1$
$\implies$ $dy$ $\,=\,$ $1 \times dz$
$\implies$ $dy$ $\,=\,$ $dz$
$\implies$ $dz$ $\,=\,$ $dy$
Now, express the first integral function in terms of $y$ by substitution.
$=\,\,\,$ $\displaystyle \int{\dfrac{1}{y}\,} dy$ $-$ $\log_e{z}+c_2$
Now, evaluate the integration of the function by the same rule.
$=\,\,\,$ $\log_e{y}+c_1$ $-$ $\log_e{z}+c_2$
$=\,\,\,$ $\log_e{y}-\log_e{z}$ $+$ $c_1+c_2$
The logarithmic expression can be simplified by the quotient rule of logarithms.
$=\,\,\,$ $\log_e{\Bigg|\dfrac{y}{z}\Bigg|}+c$
Actually, the value of $y$ is $z-1$. So, substitute it in the solution.
$=\,\,\,$ $\log_e{\Bigg|\dfrac{z-1}{z}\Bigg|}+c$
In fact, the integral of the exponential rational function is given in terms of $x$. So, we have to bring the solution back to $x$ by substituting $z = 1+e^x$.
$=\,\,\,$ $\log_e{\Bigg|\dfrac{1+e^x-1}{1+e^x}\Bigg|}+c$
$=\,\,\,$ $\require{cancel} \log_e{\Bigg|\dfrac{\cancel{1}+e^x-\cancel{1}}{1+e^x}\Bigg|}+c$
$=\,\,\,$ $\log_e{\Bigg|\dfrac{e^x}{1+e^x}\Bigg|}+c$
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