Question

Find the antiderivative of $f\left( x \right)=\ln \left( \ln x \right)+{{\left( \ln x \right)}^{-2}}$ whose graph passes through $\left( e,e \right)$

Answer 1

Antiderivative means integration and used for the opposite of the word derivative. Apply integration by parts wherever required which is given as
$\int{f\left( x \right)g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx-\int{{{f}^{'}}\left( x \right)\int{g\left( x \right)dxdx}}}$
And use the following identities
$\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x},\int{{{x}^{n}} dx=\dfrac{{{x}^{n+1}}}{n+1}}$

Complete step-by-step answer:
The word antiderivative is the substitution of the word ‘integration’ or opposite of derivative.
So, we have function $f\left( x \right)=\ln \left( \ln x \right)+{{\left( \operatorname{lnx} \right)}^{-2}}$ where we need to determine antiderivative of f (x) if it’s passing through (e, e).
Now, let I be the antiderivative of f(x); hence, we can write
$I=\int{\left( \ln \left( \ln x \right)+{{\left( \ln x \right)}^{-2}} \right)dx}$ ……………… (i)
Now, we can substitute ‘lnx’ as ‘t’ for simplifying the given integral so, we have
lnx = t or
${{\log }_{e}}x=t$
As we know that ${{\log }_{a}}x=N$ can be written as $x={{a}^{N}}$. So, we get from the above relation as:
$x={{e}^{t}}$
Now, differentiate it w.r.t ‘x’
$\begin{aligned} & \dfrac{d}{dx}\left( x \right)=\dfrac{d}{dx}\left( {{e}^{t}} \right) \\\ & 1={{e}^{t}}\dfrac{dt}{dx} \\\ & dx={{e}^{t}}dt...............\left( ii \right) \\\ \end{aligned}$
$\because \dfrac{d}{dx}{{e}^{x}}={{e}^{x}}$
So, we can re-write equation (i) with respect to variable ‘t’ as
$I=\int{\left( \ln t+{{t}^{-2}} \right)}{{e}^{t}}dt$
$\begin{aligned} & I=\int{{{e}^{t}}\operatorname{lntdt}+\int{\dfrac{{{e}^{t}}}{{{t}^{2}}}}}dt \\\ & \text{ }{{\text{I}}_{\text{1 }}}\text{ }{{\text{I}}_{\text{2}}} \\\ \end{aligned}$ …………………… (iii)
Let us calculate value of ${{I}_{2}}$ with the help of integration by parts which is given as
$\int{f\left( x \right)g\left( x \right)dx=f\left( x \right)}\int{g\left( x \right)}dx-\int{{{f}^{'}}\left( x \right)\int{g\left( x \right)dxdx}}$
So, ${{I}_{2}}$ can be expressed in form of integration by parts by supposing $f\left( x \right)={{e}^{t}}$and $g\left( x \right)=\left( \dfrac{1}{{{t}^{2}}} \right)$ in the above equation by the ILATE rule, it can be given as:
ILATE stands for
I – Inverse trigonometry
L – Logarithm
A – Algebraic
T – Trigonometry
E – Exponential
It means ILATE consists of 5 different functions as mentioned above. So, the purpose of this rule is to decide the priority of the first function and second i.e. which function is going for the integration and which is going for the differentiation. So, according to the above order, we decide the priority of function (1 and 2) involved in the integration by parts. So, we can use above mentioned identity with the given integral ${{I}_{2}}$
Hence, we get
${{I}_{2}}=\int{{{e}^{t}}{{t}^{-2}}dt={{e}^{t}}\int{{{t}^{-2}}dt-\int{\dfrac{d}{dt}{{e}^{t}}\int{{{t}^{-2}}}}}}dtdt$
As we know
$\int{{{x}^{n}}dx=\dfrac{{{x}^{n}}+1}{n+1}},\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}$
Hence, on simplifying equation of ${{I}_{2}}$ , we get
${{I}_{2}}={{e}^{t}}\left( \dfrac{-1}{t} \right)-\int{{{e}^{t}}\left( \dfrac{-1}{t} \right)dt}$
Or
${{I}_{2}}=\dfrac{-{{e}^{t}}t}{t}+\int{\dfrac{{{e}^{t}}}{t}dt}$
Now, using the integration by parts again with the $\int{\dfrac{{{e}^{t}}}{t}dt}$, we get
${{I}_{2}}=\dfrac{-{{e}^{t}}}{t}+{{e}^{t}}\int{\dfrac{1}{t}dt-\int{\dfrac{d}{dt}{{e}^{t}}\int{\dfrac{1}{t}dtdt}}}$
Now, we know
$\int{\dfrac{1}{x}dx={{\log }_{e}}x}$
$\dfrac{d}{dx}={{\log }_{e}}x$
Hence, we get
${{I}_{2}}=\dfrac{-{{e}^{t}}}{t}+{{e}^{t}}{{\log }_{e}}t-\int{{{e}^{t}}{{\log }_{e}}tdt}+c.............\left( iv \right)$
Now, we can put value of ${{I}_{2}}$ from equation (iv) to equation (iii) and hence, we get
$I=\int{{{e}^{t}}{{\log }_{e}}tdt+\dfrac{-{{e}^{t}}}{t}+{{e}^{t}}{{\log }_{e}}t+c-\int{{{e}^{t}}{{\log }_{e}}tdt}}$
So, we can cancel the term $\int{{{e}^{t}}{{\log }_{e}}tdt}$ from the above expression.
Hence, we get
$I=\dfrac{-{{e}^{t}}t}{t}+{{e}^{t}}{{\log }_{e}}t+c$
Now, substitute the value of t as ${{\log }_{e}}x$. Hence, we get
$I=\dfrac{-{{e}^{{{\log }_{{{e}^{x}}}}}}}{{{\log }_{e}}x}+{{e}^{{{\log }_{{{e}^{x}}}}}}{{\log }_{e}}{{\log }_{e}}x+c$
As, we know ${{\left( a \right)}^{{{\log }_{{{a}^{b}}}}}}=b$. Hence, we get
$I=\dfrac{-x}{{{\log }_{e}}x}+x{{\log }_{e}}\left( {{\log }_{e}}x \right)+c$
Hence, antiderivative of function f(x) is given as
$I=\dfrac{-x}{{{\log }_{e}}x}+x{{\log }_{e}}\left( {{\log }_{e}}x \right)+c$
Or
$y=\dfrac{-x}{{{\log }_{e}}x}+x{{\log }_{e}}\left( {{\log }_{e}}x \right)+c$………………. (v)
Now, we can find the value of ’c’ by using the information that the graph of antiderivative is passing through (e, e).
Hence, (e, e) will the satisfy the equation (v), so, we get
$e=\dfrac{-e}{{{\log }_{e}}e}+e{{\log }_{e}}\left( {{\log }_{e}}e \right)+c$
So, we know ${{\log }_{e}}e=1$ and ${{\log }_{e}}1=0$. Hence we get
e = - e + e (o )+ c
c = 2e
Hence, equation of antiderivative of f(x) can be given from equation (v) ad
$\dfrac{-x}{{{\log }_{e}}x}+x{{\log }_{e}}\left( {{\log }_{e}}x \right)+2e$

Note: One can integrate the term $\int{{{e}^{t}}}\ln tdt$ as well for getting the answer. But we can not find the exact values of ${{I}_{1}}$ and ${{I}_{2}}$. So, integrate only one function to cancel out the other. It is the key point for getting the integration I and hence the solution as well.
One may get confused with the term antiderivative. He or she may calculate differentiation of the given function and then try to calculate the inverse of the derivative as well, which is wrong. Antiderivative used for the word integration. So, be clear with the terms as well for these kinds of questions.