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Question: If \(n \in N\) and \({I_n} = \int {\left( {\log x} \right)^n}dx\), then \({I_n} + n{I_{n - 1}} = \) ...

If n∈Nn \in N and In=∫(log⁡x)ndx{I_n} = \int {\left( {\log x} \right)^n}dx, then In+nIn−1={I_n} + n{I_{n - 1}} =
A. (log⁡x)n+1n+1{\text{A}}{\text{. }}\dfrac{{{{\left( {\log x} \right)}^{n + 1}}}}{{n + 1}}
B. x(log⁡x)n+c{\text{B}}{\text{. }}x{\left( {\log x} \right)^n} + c
C. (log⁡x)n−1{\text{C}}{\text{. }}{\left( {\log x} \right)^{n - 1}}
D. (log⁡x)nn{\text{D}}{\text{. }}\dfrac{{{{\left( {\log x} \right)}^n}}}{n}

Explanation

Solution

Hint- Here, we will proceed by using ILATE rule of integration in order to convert the function (log⁡x)n{\left( {\log x} \right)^n} which is to be integrated in terms of function (log⁡x)n−1{\left( {\log x} \right)^{n - 1}}. Through this we will be getting an equation between In{I_n} and In−1{I_{n - 1}}.

Complete step-by-step solution -
Given, In=∫(log⁡x)ndx=∫[(log⁡x)n×1]dx {I_n} = \int {\left( {\log x} \right)^n}dx = \int \left[ {{{\left( {\log x} \right)}^n} \times 1} \right]dx{\text{ }} ………………(1)
Above integration can be carried out by using ILATE method.
According to ILATE method of integration
If we have two functions of x as u(x)u\left( x \right) and v(x)v\left( x \right)where u(x)u\left( x \right) is the first function and v(x)v\left( x \right) is the second function according to priority order of ILATE.
Then, \int u\left( x \right)v\left( x \right)dx = u\left( x \right)\int v\left( x \right)dx - \int \left\\{ {\dfrac{{d\left[ {u\left( x \right)} \right]}}{{dx}}\left[ {\int v\left( x \right)dx} \right]} \right\\}dx + c
Where c is any constant of integration.
Now in the integral given by equation (1), consider (log⁡x)n{\left( {\log x} \right)^n} as the first function and 1 as the second function.
Therefore,

{I_n} = \int \left[ {{{\left( {\log x} \right)}^n} \times 1} \right]dx{\text{ }} = {\left( {\log x} \right)^n}\int \left( 1 \right)dx - \int \left\\{ {\left[ {\dfrac{d}{{dx}}{{\left( {\log x} \right)}^n}} \right]\left[ {\int \left( 1 \right)dx} \right]} \right\\}dx + c \\\ \Rightarrow {I_n} = x{\left( {\log x} \right)^n} - \int \left\\{ {x\left[ {n{{\left( {\log x} \right)}^{n - 1}}\dfrac{{d\left( {\log x} \right)}}{{dx}}} \right]} \right\\}dx + c = x{\left( {\log x} \right)^n} - \int \left\\{ {x\left[ {n{{\left( {\log x} \right)}^{n - 1}}\left( {\dfrac{1}{x}} \right)} \right]} \right\\}dx + c \\\ \Rightarrow {I_n} = x{\left( {\log x} \right)^n} - n\int \left[ {{{\left( {\log x} \right)}^{n - 1}}} \right]dx + c{\text{ }} $$ ………… (2) Now, let us replace $n$ with $\left( {n - 1} \right)$ in equation (1) ${I_{n - 1}} = \int {\left( {\log x} \right)^{n - 1}}dx$ Substituting the value of $\int {\left( {\log x} \right)^{n - 1}}dx$ from the above equation in equation (2), we get

\Rightarrow {I_n} = x{\left( {\log x} \right)^n} - n\left( {{I_{n - 1}}} \right) + c \\
\Rightarrow {I_n} + n\left( {{I_{n - 1}}} \right) = x{\left( {\log x} \right)^n} + c \\
\\

Clearly, option B is correct. Note- In ILATE, I refer to inverse trigonometric function, L refer to logarithmic function, A refer to algebraic function, T refer to trigonometric function and E refer to exponential function. According to this priority rule the given functions are decided as the first and second functions. In this problem, ${\left( {\log x} \right)^n}$ is logarithmic function and 1 is algebraic function that’s why we have taken ${\left( {\log x} \right)^n}$ as the first function and 1 as the second function.