Question

The anti-derivative of $f\left( x \right) = \log \left( {\log x} \right) + {\left( {\log x} \right)^{ - 2}}$ whose graph passes through $\left( {e,e} \right)$ is
A. $x\left[ {\log \left( {\log x} \right) + {{\left( {\log x} \right)}^{ - 1}}} \right]$
B. $x\left[ { - \log \left( {\log x} \right) + {{\left( {\log x} \right)}^{ - 1}}} \right] + e$
C. $x\left[ {\log \left( {\log x} \right) - {{\left( {\log x} \right)}^{ - 1}}} \right] + 2e$
D. $x\left[ { - \log \left( {\log x} \right) - {{\left( {\log x} \right)}^{ - 1}}} \right] + 3e$

Answer 1

Hint: The expression of $f\left( x \right)$ is a polynomial in $\left( {\log x} \right)$, so we will substitute $\left( {\log x} \right) = t$ to simplify the above expression and then find the integral or anti-derivative of the given function. We will also use a special integral to find the anti-derivative, i.e. \[I = \int {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx} = {e^x}f\left( x \right) + C\]. The final integral will contain an arbitrary constant which can be eliminated by substituting the given point in the final expression.

Complete step by step answer:

We have, $f\left( x \right) = \log \left( {\log x} \right) + {\left( {\log x} \right)^{ - 2}}$
Anti-derivative of a function of the inverse function of derivative, or we can say anti-derivative is the integral function.
Let antiderivative function of $f\left( x \right)$is $I$
So, by the definition of antiderivative, we have:
$I = \int {f\left( x \right)dx} $
$ \Rightarrow I = \int {\left( {\log \left( {\log x} \right) + {{\left( {\log x} \right)}^{ - 2}}} \right)dx} $
As, the given function is a function of $\log x$
So, we will substitute $\left( {\log x} \right) = t$,
Now, we have
$t = \log x$
Differentiating both sides with respect to x:
\[ \Rightarrow {\text{dt = d}}\left( {{\text{logx}}} \right)\]
\[ \Rightarrow dt = \dfrac{1}{x}dx\]
\[ \Rightarrow {\text{dx = }}{{\text{e}}^{\text{t}}}{\text{dt}}\]
Now, substituting\[dx = {e^t}dt\], we have
$I = \int {\left( {\log \left( t \right) + {{\left( t \right)}^{ - 2}}} \right){e^t}dt} $
$ \Rightarrow {\text{I = }}\int {\left( {{\text{log}}\left( {\text{t}} \right){\text{ + }}\dfrac{{\text{1}}}{{{{\text{t}}^{\text{2}}}}}} \right){{\text{e}}^{\text{t}}}{\text{dt}}} $
$ \Rightarrow {\text{I = }}\int {{{\text{e}}^{\text{t}}}{\text{logtdt + }}\int {{{\text{e}}^{\text{t}}}\dfrac{{\text{1}}}{{{{\text{t}}^{\text{2}}}}}} {\text{dt}}} $
Now, adding and subtracting \[\dfrac{{\text{1}}}{{\text{t}}}\] in, \[\left( {\log \left( t \right) + \dfrac{1}{{{t^2}}}} \right)\] in order to make the above term easy for simplification and substitution.
$ \Rightarrow I = \int {{e^t}\left( {\log t + \dfrac{1}{t}} \right)dt + \int {{e^t}\left( {\dfrac{1}{{{t^2}}} - \dfrac{1}{t}} \right)} dt} $
Taking negative sign common from the second part of the above equation and simplify it as below given,
$ \Rightarrow I = \int {{e^t}\left( {\log t + \dfrac{1}{t}} \right)dt - \int {{e^t}\left( {\dfrac{1}{t} - \dfrac{1}{{{t^2}}}} \right)} dt} $
Now, we know that the following special integral,
\[I = \int {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx} = {e^x}f\left( x \right) + C\]
Thus, by using the above property, we have:
$\int {{e^t}\left( {\log t + \dfrac{1}{t}} \right)dt = } {e^t}\log t + C$ and
\[\int {{e^t}\left( {\dfrac{1}{t} - \dfrac{1}{{{t^2}}}} \right)} dt = {e^t}\dfrac{1}{t} + B\]
Where C and B are arbitrary constants.
$ \Rightarrow {\text{I = }}{{\text{e}}^{\text{t}}}{\text{logt + C - }}\dfrac{{{{\text{e}}^{\text{t}}}}}{{\text{t}}}{\text{ + B}}$
\[ \Rightarrow {\text{I = }}{{\text{e}}^{\text{t}}}\left( {{\text{logt - }}\dfrac{{\text{1}}}{{\text{t}}}} \right){\text{ + A}}\], where A is an arbitrary constant.
Now, substituting back, $t = \log x$ to find the antiderivative of the given function in terms of x, we have:
\[ \Rightarrow {\text{I = }}{{\text{e}}^{{\text{logx}}}}\left( {{\text{log}}\left( {{\text{logx}}} \right) - \dfrac{{\text{1}}}{{{\text{logx}}}}} \right){\text{ + A}}\]
\[ \Rightarrow {\text{I = x}}\left( {{\text{log}}\left( {{\text{logx}}} \right) - {{\left( {{\text{logx}}} \right)}^{{\text{ - 1}}}}} \right){\text{ + A}}\]
Now, as given, the graph of antiderivative passes through $\left( {e,e} \right)$, therefore substituting x and y both as e
\[ \Rightarrow I = x\left( {\log \left( {\log x} \right) - {{\left( {\log x} \right)}^{ - 1}}} \right) + A\]
\[ \Rightarrow e = e\left( {\log \left( {\log e} \right) - {{\left( {\log e} \right)}^{ - 1}}} \right) + A\]
\[ \Rightarrow e = e\left( {\log 1 - 1} \right) + A\]
\[ \Rightarrow 2e = A\]
\[ \Rightarrow A = 2e\]
So, we have \[A = 2e\], substituting \[A = 2e\] in the above result,
Thus, \[I = x\left( {\log \left( {\log x} \right) - {{\left( {\log x} \right)}^{ - 1}}} \right) + 2e\]
Hence, the derivative function of $f\left( x \right) = \log \left( {\log x} \right) + {\left( {\log x} \right)^{ - 2}}$ is\[x\left( {\log \left( {\log x} \right) + {{\left( {\log x} \right)}^{ - 1}}} \right) + 2e\], so option (c) is the correct answer.

Note: Students must remember the special integrals carefully and apply them in the problems to make them easy. Ability to convert a given integral into a known integral whose solution is known is also an important thing. The special integral used in the above problem is: \[I = \int {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx} = {e^x}f\left( x \right) + C\]
We will prove the above integral,
\[I = \int {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx} \]
\[I = \int {{e^x}f\left( x \right)} + \int {{e^x}f'\left( x \right)} \]
Now, using integrating by parts, using
$\int {uvdx} = u\int {vdx} - \int {\dfrac{{du}}{{dx}}\left( {\int {vdx} } \right)dx} $
So,
\[\int {{e^x}f\left( x \right)} = f\left( x \right)\int {{e^x}dx} - \int {\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}}\left( {\int {{e^x}dx} } \right)dx} \]
\[ \Rightarrow \int {{e^x}f\left( x \right)} = f\left( x \right){e^x} - \int {f'\left( x \right){e^x}dx} + C\]
\[ \Rightarrow \int {f'\left( x \right){e^x}dx} + \int {{e^x}f\left( x \right)} = f\left( x \right){e^x} + C\]
\[ \Rightarrow \int {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx} = {e^x}f\left( x \right) + C\]
Thus, \[I = \int {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx} = {e^x}f\left( x \right) + C\]
Hence, proved