# Evaluate the indefinite integral. \int \sqrt{ x^2 + 1 + 2x}dx

## Question:

Evaluate the indefinite integral.

{eq}\displaystyle \int \sqrt{ x^2 + 1 + 2x}\;dx {/eq}

## Integration by substitution:

An integral without both upper and lower limits are called indefinite integral. Apply the integration by substitution method to solve the integral. It is the one of the methods to solve the integral. It is also called a u-substitution method. While using the substitution method consider {eq}u = x + 1 {/eq}.

{eq}\text{Solution}: \\ \begin{align*} \int \sqrt{ x^2 + 1 + 2x}\;dx &= \int \sqrt{ (x + 1)^2}\;dx \\ &= \int (x + 1)^2\;dx \\ \end{align*} \\ \text{Apply the u-substitution method}: \\ . u = x + 1 \\ du = dx \\ \text{Therefore}, \\ \int (x + 1) \;dx = \int u\;du \text{Apply the power rule}: \\ \displaystyle {\color{Red}{\int x^{a}dx = \frac{x^{a+1}}{a+1}}} \\ {/eq}

{eq}\text{Therefore}, \\ \begin{align*} \int u \; du &= \frac{u^{1 + 1}}{1 + 1} \\ &= \frac{u^{2}}{2} \\ &= \frac{(x + 1)^{2}}{2} & \left ( \text{Where}, \; u = x + 1 \right ) \\ &= \frac{(x + 1)^{2}}{2} + C & \left ( \text{Add the constant term to the solution} \right ) \\ \end{align*} \\ \boxed{\text{The solution of the integral} \; {\color{Blue}{\displaystyle \int \sqrt{ x^2 + 1 + 2x}\;dx = \frac{(x + 1)^{2}}{2} + C}}} {/eq} 