# Find the solution of the integral: \int \frac{\ln x}{x} + 1 \;dx

## Question:

Find the solution of the integral:

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

## Integration by substitution:

An integral without boundaries is called indefinite integral. To find the solution of the integral first applies the sum rule. The sum rule is {eq}\displaystyle {\color{Green}{\int f(y)\pm g(y)\;dy = \int f(y)\;dy \pm \int g(y)\;dy}} {/eq}. And then apply the integration by parts to solve the integral. It is one of the methods to solve the integral. It is also called a u-substitution.

{eq}\text{Solution}: \\ \text{Given}: \\ \displaystyle \int \frac{\ln x}{x} + 1 \;dx \\ \text{Apply the sum rule}: \\ \displaystyle {\color{Red}{\int f(x)\pm g(x)\;dx = \int f(x)\;dx \pm \int g(x)\;dx}} \\ \text{Therefore}, \\ \displaystyle \int \frac{\ln x}{x} + 1 \;dx = \int \frac{\ln x}{x}\;dx + \int 1 \;dx \\ \text{Consider}, \\ \displaystyle \int \frac{\ln x}{x}\;dx \\ \text{Apply u-susbtitution}: \\ u = \ln x \\ \displaystyle du = \frac{1}{x}\;du \\ \text{Therefore}, \\ \begin{align*} \int \frac{\ln x}{x}\;dx &= \int u\;du \\ &= \frac{u^{1 + 1}}{1 + 1} & \left ( \text{Use the power rule}: \; {\color{Green}{\int x^{a}dx = \frac{x^{a+1}}{a+1} \;\; , a \neq -1}} \right ) \\ &= \frac{u^{2}}{2} \\ &= \frac{\ln^{2} x }{2} \\ &= \frac{1}{2} \ln^2 x \\ \end{align*} {/eq}

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