# Evaluate $\displaystyle \int{\dfrac{1}{1+e^x} \,} dx$

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.

$\displaystyle \int{\dfrac{1}{1+e^x} \,} dx$

### Simplify the complexity of the function

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 $x$.

$\implies$ $dx \,=\, \dfrac{dz}{z-1}$

$\implies$ $\displaystyle \int{\dfrac{1}{1+e^x} \,} dx$ $\,=\,$ $\displaystyle \int{\dfrac{1}{z} \,} \Bigg(\dfrac{dz}{z-1}\Bigg)$

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{z(z-1)} \,} dz$

### Decompose Rational expression into Partial fractions

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.

$\implies$ $\displaystyle \int{\dfrac{1}{z(z-1)} \,} dz$ $\,=\,$ $\displaystyle \int{\Bigg(\dfrac{-1}{z}+\dfrac{1}{z-1}\Bigg) \,} dz$

$\implies$ $\displaystyle \int{\dfrac{1}{z(z-1)} \,} dz$ $\,=\,$ $\displaystyle \int{\Bigg(-\dfrac{1}{z}+\dfrac{1}{z-1}\Bigg) \,} dz$

$\implies$ $\displaystyle \int{\dfrac{1}{z(z-1)} \,} dz$ $\,=\,$ $\displaystyle \int{\Bigg(\dfrac{1}{z-1}-\dfrac{1}{z}\Bigg) \,} dz$

### Evaluate the Indefinite integration of each function

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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