# Evaluate: \int \frac{x}{x + 1}\;dx

## Question:

Evaluate:

{eq}\displaystyle \int \frac{x}{x + 1}\;dx {/eq}

## Integration by substitution:

The given integral is an indefinite integral. An integral without boundaries is called indefinite integral. Integration by substitution method is one of the methods to solve the integral. It is also called a u-substitution. Consider, u = x + 1 while applying u-substitution. Apply the sum rule to integrate. The sum rule is {eq}\displaystyle {\color{Green}{\int f(x)\pm g(x) = \int f(x)dx \pm \int g(x)dx}} {/eq}.

{eq}\text{Solution}: \\ \text{Given}: \\ \displaystyle \int \frac{x}{x + 1}\;dx \\ \text{Apply u-substituton}: \\ u = x + 1 \\ du = dx \\ \text{Therefore}, \\ \begin{align*} \int \frac{x}{x + 1}\;dx &= \int \frac{u - 1}{u}\;du \\ &= \int 1 - \frac{1}{u}\;du \\ &= \int 1\;du - \int \frac{1}{u}\;du & \left ( \text{Apply the sum rule}: \; {\color{Red}{\int f(x)\pm g(x) = \int f(x)dx \pm \int g(x)dx}} \right ) \\ &= u - \ln |u| & \left ( \text{Apply the common integral} \right ) \\ &= x + 1 - \ln |x + 1| & \left ( \text{Where}, \;u = x + 1 \right ) \\ &= x + 1 - \ln |x + 1| + C & \left ( \text{Add the constant term to the solution} \right ) \\ \end{align*} \\ \boxed{\text{The solution of the integral} \; {\color{Blue}{\displaystyle \int \frac{x}{x + 1}\;dx = x + 1 - \ln |x + 1| + C}}} {/eq} 