Find $\int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}$

Answer Verified Verified
Hint: Use the formula for integration by parts,
\[\int{u(x)\cdot v(x)dx}=u(x)\cdot \int{v(x)dx}-\int{\left( \dfrac{d}{dx}u(x)\cdot \int{v(x)dx} \right)dx}\] with $u(x)=\ln x$ and $v(x)=\dfrac{1}{{{x}^{2}}}$.

Complete step-by-step answer:
First, let us decompose the function that we have to find the integration of (the integrand) into two functions.

Thus, $\int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\int\limits_{1}^{2}{\left( \ln x \right)\cdot \left( \dfrac{1}{{{x}^{2}}} \right)dx}$

We do this so that now we have two different functions and each of these functions is a simple function. Now, since we have 2 different functions, we can use integration by parts to solve the required integration. Using the ILATE rule for integration by parts, we can apply the formula\[\int{u(x)\cdot v(x)dx}=u(x)\cdot \int{v(x)dx}-\int{\left( \dfrac{d}{dx}u(x)\cdot \int{v(x)dx} \right)dx}\].

In the above formula, by the ILATE rule, the function $u(x)=\ln x$ and $v(x)=\dfrac{1}{{{x}^{2}}}$.

Using these in the formula,
\[\int{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( \ln x \right)\cdot \int{\dfrac{1}{{{x}^{2}}}dx}-\int{\left( \dfrac{d}{dx}\left( \ln x \right)\cdot \int{\dfrac{1}{{{x}^{2}}}dx} \right)dx}\ \ \ \ \ \ \ \ldots \left( 1 \right)\]
We know that \[\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}\] and we can find \[\int{\dfrac{1}{{{x}^{2}}}dx}\] as follows:
 & \Rightarrow \int{\dfrac{1}{{{x}^{2}}}dx}=\dfrac{{{x}^{-2+1}}}{-1} \\
 & \Rightarrow \int{\dfrac{1}{{{x}^{2}}}dx}=\dfrac{-1}{x} \\
Using the values of \[\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}\] and \[\int{\dfrac{1}{{{x}^{2}}}dx}=\dfrac{-1}{x}\] in the equation (1)
 & \int{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( \ln x \right)\cdot \left( \dfrac{-1}{x} \right)-\int{\left( \left( \dfrac{1}{x} \right)\cdot \left( \dfrac{-1}{x} \right) \right)dx} \\
 & \Rightarrow \int{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( \dfrac{-\ln x}{x} \right)-\int{\left( \dfrac{-1}{{{x}^{2}}} \right)dx} \\
 & \Rightarrow \int{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( \dfrac{-\ln x}{x} \right)+\int{\dfrac{1}{{{x}^{2}}}dx}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 2 \right) \\
Using the value of \[\int{\dfrac{1}{{{x}^{2}}}dx}=\dfrac{-1}{x}\] in equation (2),
 & \int{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( \dfrac{-\ln x}{x} \right)+\left( \dfrac{-1}{x} \right)\ \\
 & \Rightarrow \int{\dfrac{\ln x}{{{x}^{2}}}dx}=-\dfrac{\ln x}{x}-\dfrac{1}{x}+C \\
Putting in the lower and upper limit in the integration obtained,
 & \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( -\dfrac{\ln x}{x}-\dfrac{1}{x} \right)_{1}^{2} \\
 & \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( -\dfrac{\ln 2}{2}-\dfrac{1}{2} \right)-\left( -\dfrac{\ln 1}{1}-\dfrac{1}{1} \right) \\
 & \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( -\dfrac{\ln 2}{2}-\dfrac{1}{2} \right)-\left( 0-1 \right) \\
 & \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( -\dfrac{\ln 2}{2}-\dfrac{1}{2} \right)-\left( -1 \right) \\
 & \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=-\dfrac{\ln 2}{2}-\dfrac{1}{2}+1 \\
 & \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=-\dfrac{\ln 2}{2}+\dfrac{1}{2} \\

 & \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\dfrac{1}{2}-\dfrac{\ln 2}{2} \\
 & \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\dfrac{1-\ln 2}{2} \\
Thus the required integration of $\int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}$ is \[\dfrac{1-\ln 2}{2}\].

Note: At first sight of the question, a student may be tempted to use integration by substitution method and apply $\ln x=t$ because the differentiation of $\ln x=\dfrac{1}{x}$ is there in the question. But because of the square attached, this approach fails to give the correct answer. Had the question been $\int\limits_{1}^{2}{\dfrac{\ln x}{x}dx}$, in that case, such an approach would have provided the correct results but because of ${{x}^{2}}$ in the denominator, integration by parts is the more suitable approach for finding the correct answer.
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