# Evaluate the indefinite integral: \int \frac{1}{t + 1}\;dt

## Question:

Evaluate the indefinite integral:

{eq}\displaystyle \int \frac{1}{t + 1}\;dt {/eq}

## Integration by substitution:

An integral without boundary is called indefinite integral. Integration by substitution method is one of the methods to solve the integral. It is also called the u-substitution method. While applying integration by substitution method consider {eq}u = t + 1 {/eq}.

{eq}\text{Solution}: \\ \text{Given}: \\ \displaystyle \int \frac{1}{t + 1}\;dt \\ \text{Apply integration by substitution method}: \\ u = t + 1 \\ du = dt \\ \text{Therefore}, \\ \begin{align*} \int \frac{1}{t + 1}\;dt &= \int \frac{1}{u}\;du \\ &= \ln |u| & \left ( \text{Use the common integral} \right ) \\ &= \ln |t + 1| & \left ( \text{Where}, \; u = t + 1 \right ) \\ &= \ln | t + 1 | + 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{1}{t + 1}\;dt = \ln | t + 1 | + C }}} {/eq} 