Question

If $n \in N$ and ${I_n} = \int {\left( {\log x} \right)^n}dx$, then ${I_n} + n{I_{n - 1}} = $
${\text{A}}{\text{. }}\dfrac{{{{\left( {\log x} \right)}^{n + 1}}}}{{n + 1}}$
${\text{B}}{\text{. }}x{\left( {\log x} \right)^n} + c$
${\text{C}}{\text{. }}{\left( {\log x} \right)^{n - 1}}$
${\text{D}}{\text{. }}\dfrac{{{{\left( {\log x} \right)}^n}}}{n}$

Answer 1

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

Complete step-by-step solution -
Given, ${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 $u\left( x \right)$ and $v\left( x \right)$where $u\left( x \right)$ is the first function and $v\left( x \right)$ 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 ${\left( {\log x} \right)^n}$ 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.