How do you integrate $\int{\dfrac{\ln x}{{{x}^{2}}}dx}$ by using integration by parts method?

Hint: In this problem we need to integrate the given function by using the integration by parts method. The integration by parts method say that $\int{uvdx}=u\int{vdx}-\int{\left( \left( {{u}^{'}} \right)\int{vdx} \right)dx}$. Integration by parts rule is applicable when we have two functions that are in multiplication. So, we will convert our given equation in the required form. In integration by parts rule we need to select the functions $u$, $v$ from the given equation so that they follow the ILATE rule. ILATE rule is the rule which gives the order of importance for selecting the function in integration by parts rule. According to this rule the order of selecting a function is Inverse, Logarithmic, Algebraic, Trigonometric, Exponential. By using this rule, we will select the function $u$, $v$ from the given function and we will apply the by parts formula and simplify it to get the required result.

Complete step by step answer:
Given that, $\int{\dfrac{\ln x}{{{x}^{2}}}dx}$
Writing the ${{x}^{2}}$ which is in denominator as $\dfrac{1}{{{x}^{2}}}$ in the above equation, then we will get
$\int{\dfrac{\ln x}{{{x}^{2}}}dx}=\int{\ln x\times \dfrac{1}{{{x}^{2}}}dx}$
In the above equation we can observe that the both functions are in multiplication, so we can freely use integration by parts rule.
The functions we have in the above equation are the first one is Logarithmic; Second one is Algebraic. According to ILATE rule we can choose logarithmic function as $u$ and the reaming algebraic function as $v$, then we will get
$u=\ln x$, $v=\dfrac{1}{{{x}^{2}}}={{x}^{-2}}$
Applying the integration by parts rule, then we will have
$\int{\dfrac{\ln x}{{{x}^{2}}}dx}=\ln x\int{{{x}^{-2}}dx}-\int{\left( {{\left( \ln x \right)}^{'}}\int{{{x}^{-2}}dx} \right)dx}$
Applying the known formulas $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C$, $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$ in the above equation, then we will get
$\begin{align}
  & \Rightarrow \int{\dfrac{\ln x}{{{x}^{2}}}dx}=\ln x\left( -\dfrac{1}{x} \right)-\int{\dfrac{1}{x}\times \dfrac{-1}{x}dx} \\
 & \Rightarrow \int{\dfrac{\ln x}{{{x}^{2}}}dx}=-\dfrac{\ln x}{x}+\int{{{x}^{-2}}dx} \\
\end{align}$
Again, using the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C$ in the above equation, then we will have
$\begin{align}
  & \Rightarrow \int{\dfrac{\ln x}{{{x}^{2}}}dx}=-\dfrac{\ln x}{x}+\left( -\dfrac{1}{x} \right)+C \\
 & \Rightarrow \int{\dfrac{\ln x}{{{x}^{2}}}dx}=-\dfrac{\ln x}{x}-\dfrac{1}{x}+C \\
 & \Rightarrow \int{\dfrac{\ln x}{{{x}^{2}}}dx}=-\dfrac{1}{x}\left( \ln x+1 \right)+C \\
\end{align}$

Note: In this problem we have used the integration by parts rule as $\int{uvdx}=u\int{vdx}-\int{\left( \left( {{u}^{'}} \right)\int{vdx} \right)dx}$. In some text books it can be written in different forms for example $\int{\left( u \right)\left( \dfrac{dv}{dx} \right)dx}=\left( u \right)\left( v \right)-\int{\left( v \right)\left( \dfrac{du}{dx} \right)dx}$. This form is also known as integration by part rules. Likewise, we have several formats for the integration by parts. The one which we have used is the simple one among the all-other formats. So, we have used the formula $\int{uvdx}=u\int{vdx}-\int{\left( \left( {{u}^{'}} \right)\int{vdx} \right)dx}$ and calculated the result.