How do you integrate \[\left( {\dfrac{1}{{{e^x} + 1}}} \right)dx\]?

Hint: We use the substitution of the exponential term and convert the complete equation in terms of the new variable. Then use integration by parts to solve for the value of integration.

Complete step-by-step answer:
We have to solve for the value of \[\int {\left( {\dfrac{1}{{{e^x} + 1}}} \right)dx} \] … (1)
Let us substitute the value of \[{e^x} = t\]
Then differentiating both sides of the equation we get
\[ \Rightarrow {e^x}dx = dt\]
Shift all values except dx to right side of the equation
 \[ \Rightarrow dx = \dfrac{{dt}}{{{e^x}}}\]
Put the value of \[{e^x} = t\]in RHS of the equation
\[ \Rightarrow dx = \dfrac{{dt}}{t}\] … (2)
Now substitute the value of \[{e^x} = t\] and the value of dx from equation (2) in equation (1)
\[ \Rightarrow \int {\left( {\dfrac{1}{{{e^x} + 1}}} \right)dx} = \int {\dfrac{1}{t}\left( {\dfrac{1}{{t + 1}}} \right)dt} \]
Now we know the term that has to be integrated can be broken using by parts method.
We write the integrand as sum of two integrands and solve for their respective coefficients
\[ \Rightarrow \dfrac{1}{t}\left( {\dfrac{1}{{t + 1}}} \right) = \dfrac{A}{t} + \dfrac{B}{{t + 1}}\] … (3)
Take LCM on both sides of the equation
\[ \Rightarrow \dfrac{1}{{t(t + 1)}} = \dfrac{{A(t + 1) + Bt}}{{t(t + 1)}}\]
Cancel same denominators from both sides of the equation
\[ \Rightarrow 1 = A(t + 1) + Bt\]
Now when we put the value of \[t = - 1\]we get
\[ \Rightarrow 1 = A( - 1 + 1) + B( - 1)\]
\[ \Rightarrow 1 = A \times (0) + B( - 1)\]
\[ \Rightarrow 1 = B( - 1)\]
Multiply both sides by -1
\[ \Rightarrow - 1 = B\] … (4)
Now when we put the value of \[t = 0\]we get
\[ \Rightarrow 1 = A(0 + 1) + B(0)\]
\[ \Rightarrow 1 = A \times (1)\]
\[ \Rightarrow 1 = A\] … (5)
Substitute the values of A and B in equation (3)
\[ \Rightarrow \dfrac{1}{t}\left( {\dfrac{1}{{t + 1}}} \right) = \dfrac{1}{t} + \dfrac{{ - 1}}{{t + 1}}\]
\[ \Rightarrow \dfrac{1}{t}\left( {\dfrac{1}{{t + 1}}} \right) = \dfrac{1}{t} - \dfrac{1}{{t + 1}}\]
Then the integration becomes \[\int {\dfrac{1}{t}\left( {\dfrac{1}{{t + 1}}} \right)} dt = \int {\dfrac{1}{t}dt - \int {\dfrac{1}{{t + 1}}} } dt\]
Since we know \[\int {\dfrac{1}{x}dx = \log x} + C\], we can write the integration as
\[ \Rightarrow \int {\dfrac{1}{t}dt - \int {\dfrac{1}{{t + 1}}} } dt = \log t - \log (1 + t) + C\]
We know property of log that \[\log m - \log n = \log \dfrac{m}{n}\]
\[ \Rightarrow \int {\dfrac{1}{t}dt - \int {\dfrac{1}{{t + 1}}} } dt = \log \dfrac{t}{{1 + t}} + C\]
Substitute the value of \[{e^x} = t\]
\[ \Rightarrow \int {\dfrac{1}{t}dt - \int {\dfrac{1}{{t + 1}}} } dt = \log \dfrac{{{e^x}}}{{1 + {e^x}}} + C\]

\[\therefore \]Integration of \[\left( {\dfrac{1}{{{e^x} + 1}}} \right)dx\] is \[\log \dfrac{{{e^x}}}{{1 + {e^x}}} + C\]

Note:
Many students make the mistake of cancelling the log function from the exponential function in the last step which is wrong as the log function is applied on the complete fraction, so to open it we will have to open the fraction first and then cancel it which is of no use.