# The following integral requires a preliminary step such as long division or a change of variables...

## Question:

The following integral requires a preliminary step such as long division or a change of variables before using partial fractions. Evaluate the integral {eq}\int \frac{\mathrm{d}x}{1 + e^x} {/eq}.

We need to evaluate the integral given by {eq}I = \int {\dfrac{{dx}}{{1 + {e^x}}}} {/eq}

We will be using the method of substitution to evaluate the following integral.

Let {eq}e^{x} = t {/eq}, therefore we have {eq}e^{x} dx = dt {/eq}, this means {eq}dx= \dfrac{dt}{e^x} = \dfrac{dt}{t} {/eq}.

Hence the integral is converted into

{eq}I = \int {\dfrac{{dt}}{{t\left( {t + 1} \right)}}} {/eq}

Now {eq}\dfrac{1}{ t \left( t+1 \right)} {/eq} is a proper fraction where the degree of the numerator is less than the degree of the denominator.

Therefore, this function can be factorized into sum of other functions with simpler denominators.

Since the factors of the denominator are linear and nonrepeated, therefore we have

{eq}\dfrac{1}{{t\left( {t + 1} \right)}} = \dfrac{A}{t} + \dfrac{B}{{t + 1}} {/eq}

This relation is true for all values of {eq}t {/eq}.

Therefore finding the value of {eq}t {/eq}, we have

{eq}\eqalign{ & \dfrac{1}{{t\left( {t + 1} \right)}} = \dfrac{{A\left( {t + 1} \right) + B\left( t \right)}}{{t\left( {t + 1} \right)}} \cr & \Rightarrow 1 = A\left( {t + 1} \right) + B\left( t \right) \cr & {\text{For }}t = - 1 \cr & 1 = - B \cr & {\text{For }}t = 0 \cr & 1 = A \cr} {/eq}

Therefore

{eq}\dfrac{1}{{t\left( {t + 1} \right)}} = \dfrac{1}{t} - \dfrac{1}{{t + 1}} {/eq}

which implies

{eq}I = \int {\dfrac{{dt}}{{t\left( {t + 1} \right)}}} = \ln \left| t \right| - \ln \left| {t + 1} \right| = \ln \left| {\dfrac{t}{{t + 1}}} \right| + C {/eq}

Putting back the value of {eq}t {/eq}, we get

{eq}I = \int {\dfrac{{dx}}{{1 + {e^x}}} = \color{red}{\ln \left| {\dfrac{{{e^x}}}{{{e^x} + 1}}} \right| + C}} {/eq} 