# Evaluate the integral: \int \frac{1 - \ln x}{x} \;dx

## Question:

Evaluate the integral:

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

## Integration by substitution:

An integral with both upper and lower limits are called indefinite integral. To find the solution of the integral applies the integration by substitution method. It is also a u-substitution. Consider, {eq}u = 1 - \ln (x) {/eq} while applying integration by substitution method.

{eq}\text{Solution}: \\ \text{Given}: \\ \displaystyle \int \frac{1 - \ln x}{x} \;dx \\ \text{Apply u-substitution method}: \\ u = 1 - \ln x \\ du = \frac{-1}{x} \;dx \\ \text{Therefore}, \\ \begin{align*} \int \frac{1 - \ln x}{x} \;dx &= \int - u\;du \\ &= - \int i\;du & \left ( \text{Take the constant out} \right ) \\ &= - \frac{u^{1 + 1}}{1 + 1} & \left ( \text{Apply the power rule}: {\color{Red}{\int x^{a}dx = \frac{x^{a+1}}{a+1} \;\; , a \neq -1}} \right ) \\ &= - \frac{u^{2}}{2} \\ &= - \frac{(1 - \ln x)^{2}}{2} & \left ( \text{Where}, \; u = 1 - \ln x \right ) \\ &= - \frac{1}{2}(1 - \ln x )^2 \\ &= - \frac{1}{2}(1 - \ln x )^2 + 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 - \ln x}{x} \;dx = - \frac{1}{2}(1 - \ln x )^2 + C }}} {/eq} 