Find $\int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}$
Answer
328.5k+ views
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:
\[\int{\dfrac{1}{{{x}^{2}}}dx}=\int{{{x}^{-2}}dx}\]
\[\begin{align}
& \Rightarrow \int{\dfrac{1}{{{x}^{2}}}dx}=\dfrac{{{x}^{-2+1}}}{-1} \\
& \Rightarrow \int{\dfrac{1}{{{x}^{2}}}dx}=\dfrac{-1}{x} \\
\end{align}\]
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)
\[\begin{align}
& \int{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( \ln x \right)\cdot \left( \dfrac{-1}{x} \right)-\int{\left( \left( \dfrac{1}{x} \right)\cdot \left( \dfrac{-1}{x} \right) \right)dx} \\
& \Rightarrow \int{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( \dfrac{-\ln x}{x} \right)-\int{\left( \dfrac{-1}{{{x}^{2}}} \right)dx} \\
& \Rightarrow \int{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( \dfrac{-\ln x}{x} \right)+\int{\dfrac{1}{{{x}^{2}}}dx}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 2 \right) \\
\end{align}\]
Using the value of \[\int{\dfrac{1}{{{x}^{2}}}dx}=\dfrac{-1}{x}\] in equation (2),
\[\begin{align}
& \int{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( \dfrac{-\ln x}{x} \right)+\left( \dfrac{-1}{x} \right)\ \\
& \Rightarrow \int{\dfrac{\ln x}{{{x}^{2}}}dx}=-\dfrac{\ln x}{x}-\dfrac{1}{x}+C \\
\end{align}\]
Putting in the lower and upper limit in the integration obtained,
\[\begin{align}
& \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( -\dfrac{\ln x}{x}-\dfrac{1}{x} \right)_{1}^{2} \\
& \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( -\dfrac{\ln 2}{2}-\dfrac{1}{2} \right)-\left( -\dfrac{\ln 1}{1}-\dfrac{1}{1} \right) \\
& \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( -\dfrac{\ln 2}{2}-\dfrac{1}{2} \right)-\left( 0-1 \right) \\
& \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( -\dfrac{\ln 2}{2}-\dfrac{1}{2} \right)-\left( -1 \right) \\
& \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=-\dfrac{\ln 2}{2}-\dfrac{1}{2}+1 \\
& \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=-\dfrac{\ln 2}{2}+\dfrac{1}{2} \\
\end{align}\]
\[\begin{align}
& \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\dfrac{1}{2}-\dfrac{\ln 2}{2} \\
& \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\dfrac{1-\ln 2}{2} \\
\end{align}\]
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.
\[\int{u(x)\cdot v(x)dx}=u(x)\cdot \int{v(x)dx}-\int{\left( \dfrac{d}{dx}u(x)\cdot \int{v(x)dx} \right)dx}\] 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:
\[\int{\dfrac{1}{{{x}^{2}}}dx}=\int{{{x}^{-2}}dx}\]
\[\begin{align}
& \Rightarrow \int{\dfrac{1}{{{x}^{2}}}dx}=\dfrac{{{x}^{-2+1}}}{-1} \\
& \Rightarrow \int{\dfrac{1}{{{x}^{2}}}dx}=\dfrac{-1}{x} \\
\end{align}\]
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)
\[\begin{align}
& \int{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( \ln x \right)\cdot \left( \dfrac{-1}{x} \right)-\int{\left( \left( \dfrac{1}{x} \right)\cdot \left( \dfrac{-1}{x} \right) \right)dx} \\
& \Rightarrow \int{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( \dfrac{-\ln x}{x} \right)-\int{\left( \dfrac{-1}{{{x}^{2}}} \right)dx} \\
& \Rightarrow \int{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( \dfrac{-\ln x}{x} \right)+\int{\dfrac{1}{{{x}^{2}}}dx}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 2 \right) \\
\end{align}\]
Using the value of \[\int{\dfrac{1}{{{x}^{2}}}dx}=\dfrac{-1}{x}\] in equation (2),
\[\begin{align}
& \int{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( \dfrac{-\ln x}{x} \right)+\left( \dfrac{-1}{x} \right)\ \\
& \Rightarrow \int{\dfrac{\ln x}{{{x}^{2}}}dx}=-\dfrac{\ln x}{x}-\dfrac{1}{x}+C \\
\end{align}\]
Putting in the lower and upper limit in the integration obtained,
\[\begin{align}
& \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( -\dfrac{\ln x}{x}-\dfrac{1}{x} \right)_{1}^{2} \\
& \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( -\dfrac{\ln 2}{2}-\dfrac{1}{2} \right)-\left( -\dfrac{\ln 1}{1}-\dfrac{1}{1} \right) \\
& \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( -\dfrac{\ln 2}{2}-\dfrac{1}{2} \right)-\left( 0-1 \right) \\
& \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\left( -\dfrac{\ln 2}{2}-\dfrac{1}{2} \right)-\left( -1 \right) \\
& \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=-\dfrac{\ln 2}{2}-\dfrac{1}{2}+1 \\
& \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=-\dfrac{\ln 2}{2}+\dfrac{1}{2} \\
\end{align}\]
\[\begin{align}
& \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\dfrac{1}{2}-\dfrac{\ln 2}{2} \\
& \Rightarrow \int\limits_{1}^{2}{\dfrac{\ln x}{{{x}^{2}}}dx}=\dfrac{1-\ln 2}{2} \\
\end{align}\]
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.
Last updated date: 04th Jun 2023
•
Total views: 328.5k
•
Views today: 6.87k
Recently Updated Pages
If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts
What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?
