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Question: Evaluate the following given expression: \(\int_{{}}^{{}}{\dfrac{\log x}{{{x}^{2}}}dx}\)...

Evaluate the following given expression:
∫log⁡xx2dx\int_{{}}^{{}}{\dfrac{\log x}{{{x}^{2}}}dx}

Explanation

Solution

Hint: In this problem we will apply the method of integration by parts. The method of integration by parts is generally used when we have to integrate the product of two functions. As per this method we take the functions as u and v and then proceed. We use the ILATE rule to choose u and v.

Complete step-by-step answer:

Integration by parts or partial integration is a method that finds the integral of a product of two functions in terms of the integral of the product of their derivatives and antiderivatives. This method is frequently used to transform the anti derivative of a product of two functions into an antiderivative for which a solution can be more easily found.
If we take these two functions as u and v and we have to find the value of ∫u.vdx\int_{{}}^{{}}{u.vdx}, then according to the formula of integration by parts, we have:
∫u.vdx=u.∫v.dx−∫d(u)dx.∫v.dx....................(1)\int_{{}}^{{}}{u.vdx}=u.\int_{{}}^{{}}{v.dx}-\int_{{}}^{{}}{\dfrac{d\left( u \right)}{dx}.\int_{{}}^{{}}{v.dx....................(1)}}
Since, the function given to us to integrate is ∫log⁡xx2dx\int_{{}}^{{}}{\dfrac{\log x}{{{x}^{2}}}dx} with respect to us.
Here, we will take u = logx and v=1x2v=\dfrac{1}{{{x}^{2}}} and we will substitute them in equation (1). Then we get:
∫log⁡xx2dx=log⁡x∫1x2dx−∫d(log⁡x)dx.∫1x2dx...............(2)\int_{{}}^{{}}{\dfrac{\log x}{{{x}^{2}}}dx=\log x\int_{{}}^{{}}{\dfrac{1}{{{x}^{2}}}dx}-\int_{{}}^{{}}{\dfrac{d\left( \log x \right)}{dx}.\int_{{}}^{{}}{\dfrac{1}{{{x}^{2}}}dx}}}...............\left( 2 \right)
Since, we know that ∫1x2dx=−1x   and d(log⁡x)dx=1x\int_{{}}^{{}}{\dfrac{1}{{{x}^{2}}}dx}=\dfrac{-1}{x}\,\,\,and\,\dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x}
On substituting these values in equation (2), we get:
∫log⁡xx2dx=log⁡x.(−1)x−∫1x.(−1)xdx  =−log⁡xx+∫1x2dx  =−log⁡xx+(−1)x+c \begin{aligned} & \int_{{}}^{{}}{\dfrac{\log x}{{{x}^{2}}}dx=\log x.\dfrac{\left( -1 \right)}{x}-\int_{{}}^{{}}{\dfrac{1}{x}.\dfrac{\left( -1 \right)}{x}dx}} \\\ & \,=\dfrac{-\log x}{x}+\int_{{}}^{{}}{\dfrac{1}{{{x}^{2}}}dx} \\\ & \,=\dfrac{-\log x}{x}+\dfrac{\left( -1 \right)}{x}+c \\\ \end{aligned}
Here, c is the constant of integration.
Hence, integration of the function log⁡xx2\dfrac{\log x}{{{x}^{2}}} is = −log⁡xx−1x+c\dfrac{-\log x}{x}-\dfrac{1}{x}+c

Note: While choosing u and v functions, students should keep the ILATE rule in mind. In ILATE, as we know that I stands for inverse logarithmic function, L stands for logarithmic function, A stands for algebraic function, T stands for trigonometric function and A stands for algebraic function. Here L comes before A, hence we give first preference to L.