# Find the solution of the integral: \int \frac{e^y}{1 + e^y}\;dy

## Question:

Find the solution of the integral:

{eq}\displaystyle \int \frac{e^y}{1 + e^y}\;dy {/eq}

## Integration by substitution:

An integral without both upper and lower limits are called indefinite integral. To find the solution of the integral use of the integration by substitution method. It is also called a u-substitution. Consider, {eq}u = 1 + e^y {/eq} while applying the u-substitution method.

{eq}\text{Solution}: \\ \text{Given}: \\ \displaystyle \int \frac{e^y}{1 + e^y}\;dy \\ \text{Apply u-substitution}: \\ u = 1 + e^y \\ du = e^y\;dy \\ \text{Therefore}, \\ \begin{align*} \int \frac{e^y}{1 + e^y}\;dy &= \int \frac{1}{u}\;du \\ &= \ln |u| & \left ( \text{Use the common integral} \right ) \\ &= \ln |1 + e^y| & \left ( \text{Where}, \; u = 1 + e^y \right ) \\ &= \ln |1 + e^y| + C & \left ( \text{Add the constant term to the solution} \right ) \\ \end{align*} \\ \boxed{\text{The solution of the integral} \; \displaystyle {\color{Blue}{\int \frac{e^y}{1 + e^y}\;dy = \ln |1 + e^y| + C }}} {/eq}