Question

If $\int {{{\left( {\log x} \right)}^2}dx} = x.{\left( {f\left( x \right)} \right)^2} + Ax\left[ {f\left( x \right) - 1} \right] + c$ holds then choose the correct option.
1) $f\left( x \right) = \log x,A = 2$
2) $f\left( x \right) = \log x,A = - 2$
3) $f\left( x \right) = - \log x,A = 2$
4) $f\left( x \right) = - \log x,A = - 2$

Answer 1

Hint: This is a different type of problem but there is no change in the processes of solving this problem. It is a similar way to finding $\int {\log xdx} $ . We first write the function into a function multiplied by one and expand it using the $u.v$ formula and then comparing it with the given function to get the correct option.

Complete step-by-step answer:
At first, let us find the value of $\int {{{\left( {\log x} \right)}^2}dx} $
We can write it as
$ \Rightarrow \int {{{\left( {\log x} \right)}^2}dx} = \int {{{\left( {\log x} \right)}^2}.1dx} $,
Now let us use the $u.v$ formula to the above equation, we get
$ \Rightarrow \int {{{\left( {\log x} \right)}^2}dx} = {\left( {\log x} \right)^2}.\int {1.dx - \int {\dfrac{d}{{dx}}{{\left( {\log x} \right)}^2}.\left( {\int {1.dx} } \right).} dx} $
We know that $\int {1.dx} = x$
By substituting it in the above equation we get,
$ \Rightarrow \int {{{\left( {\log x} \right)}^2}dx} = {\left( {\log x} \right)^2}.x - \int {\dfrac{d}{{dx}}{{\left( {\log x} \right)}^2}.x.} dx$
And we can have the $\dfrac{d}{{dx}}{\left( {\log x} \right)^2}$value using the chain rule
So, after using the chain rule we get,
$\dfrac{d}{{dx}}{\left( {\log x} \right)^2} = 2.\dfrac{{\log x}}{x}$.
By substituting the above value in the equation we get.
$ \Rightarrow \int {{{\left( {\log x} \right)}^2}dx} = x.{\left( {\log x} \right)^2} - \int {2.\dfrac{{\log x}}{x}.x.} dx$
We can cancel out the $x$in the second term.
After canceling it and getting two outside, we get
$ \Rightarrow \int {{{\left( {\log x} \right)}^2}dx} = x.{\left( {\log x} \right)^2} - 2.\int {\log x.} dx$,
Now we need to find the value of $\int {\log xdx} $
This is also done by the same method
$\int {\log xdx} = \int {\log x.1.dx} $,
$$ \Rightarrow \int {\log xdx} = \log x.\int {1.dx} - \int {\dfrac{d}{{dx}}\left( {\log x} \right).\int {1.dx} .dx} $$ ,
After substituting known values, we get
$$ \Rightarrow \int {\log xdx} = \log x.x - \int {\dfrac{1}{x}.x.dx} $$
$$ \Rightarrow \int {\log xdx} = x\log x - \int {1.dx} $$
$$ \Rightarrow \int {\log xdx} = x\log x - x$$
Now let us substitute the above value in the equation.
After substituting we get,
$ \Rightarrow \int {{{\left( {\log x} \right)}^2}dx} = x.{\left( {\log x} \right)^2} - 2\left( {x\log x - x} \right)$
We can take $x$ common from the second term
After taking common we get,
$ \Rightarrow \int {{{\left( {\log x} \right)}^2}dx} = x.{\left( {\log x} \right)^2} - 2.x.\left( {\log x - 1} \right) + c$ .
But given,
$\int {{{\left( {\log x} \right)}^2}dx} = x.{\left( {f\left( x \right)} \right)^2} + Ax\left[ {f\left( x \right) - 1} \right] + c$ ,
$ \Rightarrow x.{\left( {f\left( x \right)} \right)^2} + Ax\left[ {f\left( x \right) - 1} \right] + c = x.{\left( {\log x} \right)^2} - 2.x.\left( {\log x - 1} \right)$.
Therefore by comparing on both sides we get,
$f\left( x \right) = \log x,A = - 2$.
So, the correct option is 2.
So, the correct answer is “Option 2”.

Note: This is a type of problem that everyone should practice because it is tough to get an idea that we should apply the $u.v$ formula of integration by taking other functions as one. If we practiced this type of problem earlier then it is easy to get this idea in the time of the exam.