Question

Evaluate the value of integral:
$\int{\dfrac{\log x}{x}dx}$

Answer 1

Hint: For solving this problem, we use the method of substitution for integration. We let log x to some other variable t. Now, by rewriting dx in the form of dt, convert the integral in simplified form. By using the standard formula of integral, we obtain the result.
Complete step-by-step answer:
According to the problem statement, we are required to evaluate the integral: $\int{\dfrac{\log x}{x}dx}$
This integral is difficult to evaluate in terms of x. So, we use the technique of substitution to convert the variable x into some other reducible variable t which simplifies the integral. For the above integral suitable substitution is: $\log x=t\ldots (1)$
Now we have to transform dx into dt. For doing so, differentiating the equation (1) with respect to x, we get
$\Rightarrow \dfrac{d}{dx}\left( \log x \right)=\dfrac{d}{dx}\left( t \right)$
Some useful standard formulas of differentiation can be stated as: $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}\text{ and }\dfrac{d}{dx}\left( a \right)=\dfrac{da}{dx}$
$\begin{align}
  & \Rightarrow \dfrac{1}{x}=\dfrac{dt}{dx} \\
 & \Rightarrow dt=\dfrac{dx}{x}\ldots \left( 2 \right) \\
\end{align}$
Substituting the values of equation (1) and equation (2) in the initial integral, we get
$\Rightarrow \int{\dfrac{\log }{x}dx=\int{tdt}}$
This integral is easy to solve by using standard formula which can be stated as: $\int{xdx}=\dfrac{{{x}^{2}}}{2}+c$
By using the above, the value of integral is $\int{tdt=\dfrac{{{t}^{2}}}{2}+c}$
Replacing t in terms of x we get the result as $\dfrac{{{\left( \log x \right)}^{2}}}{2}+c$.
Note: Students must be familiar with the topic of integration using substitution for solving this problem. The variables present in the integral can only be used for substitution. For this case, we can substitute $\dfrac{1}{x}=t\text{ or }\log x=t$. By using the second substitution, we obtained the simplified form.