Question

Evaluate the following indefinite integral \[\int{\dfrac{{{x}^{2}}\left( 1-\ln x \right)}{{{\ln }^{4}}x-{{x}^{4}}}}dx\]
(a) $\dfrac{1}{2}\ln \left( \dfrac{x}{\ln x} \right)-\dfrac{1}{4}\ln \left( {{\ln }^{2}}x-{{x}^{2}} \right)+C$,
(b) $\dfrac{1}{4}\ln \left( \dfrac{\ln \left( x \right)-x}{\ln \left( x \right)+x} \right)-\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{\ln x}{x} \right)+C$,
(c) $\dfrac{1}{4}\ln \left( \dfrac{\ln \left( x \right)+x}{\ln \left( x \right)-x} \right)-\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{\ln x}{x} \right)+C$,
(d) $\dfrac{1}{4}\left( \ln \left( \dfrac{\ln \left( x \right)-x}{\ln \left( x \right)+x} \right)+{{\tan }^{-1}}\left( \dfrac{\ln x}{x} \right) \right)+C$.

Answer 1

Hint: We start solving the problem by assigning a variable to the given integral. We then make the necessary arrangements required in integrand to divide it into the sum of two integrals. We then calculate each integral separately by assuming $\dfrac{\ln x}{x}=t$ in each integral. We then use the integration formulas $\int{\dfrac{dx}{{{x}^{2}}-{{a}^{2}}}}=\dfrac{1}{2a}\ln \left( \dfrac{x-a}{x+a} \right)+C$ and $\int{\dfrac{dx}{{{x}^{2}}+{{a}^{2}}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)+C$ to find the integrals in terms of ‘t’. We then substitute $\dfrac{\ln x}{x}$ in place of t to get the required result.

Complete step-by-step answer:
According to the problem, we need to solve the indefinite integral \[\int{\dfrac{{{x}^{2}}\left( 1-\ln x \right)}{{{\ln }^{4}}x-{{x}^{4}}}}dx\]. Let us assume the integral be I.
So, we have $I=\int{\dfrac{{{x}^{2}}\left( 1-\ln x \right)}{{{\ln }^{4}}x-{{x}^{4}}}}dx$ ----(1).
Let us multiply and divide integrand with 2.
$\Rightarrow I=\int{\dfrac{2{{x}^{2}}\left( 1-\ln x \right)}{2\left( {{\ln }^{4}}x-{{x}^{4}} \right)}}dx$.
$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{2{{x}^{2}}\left( 1-\ln x \right)}{\left( {{\ln }^{2}}x-{{x}^{2}} \right)\left( {{\ln }^{2}}x+{{x}^{2}} \right)}}dx$.
$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( 2{{x}^{2}}+{{\ln }^{2}}x-{{\ln }^{2}}x \right)\left( 1-\ln x \right)}{\left( {{\ln }^{2}}x-{{x}^{2}} \right)\left( {{\ln }^{2}}x+{{x}^{2}} \right)}}dx$.
$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( {{\ln }^{2}}x+{{x}^{2}}-{{\ln }^{2}}x+{{x}^{2}} \right)\left( 1-\ln x \right)}{\left( {{\ln }^{2}}x-{{x}^{2}} \right)\left( {{\ln }^{2}}x+{{x}^{2}} \right)}}dx$.
$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( \left( {{\ln }^{2}}x+{{x}^{2}} \right)-\left( {{\ln }^{2}}x-{{x}^{2}} \right) \right)\left( 1-\ln x \right)}{\left( {{\ln }^{2}}x-{{x}^{2}} \right)\left( {{\ln }^{2}}x+{{x}^{2}} \right)}}dx$.
\[\Rightarrow I=\dfrac{1}{2}\int{\left( \dfrac{\left( {{\ln }^{2}}x+{{x}^{2}} \right)\left( 1-\ln x \right)}{\left( {{\ln }^{2}}x-{{x}^{2}} \right)\left( {{\ln }^{2}}x+{{x}^{2}} \right)} \right)-\left( \dfrac{\left( {{\ln }^{2}}x-{{x}^{2}} \right)\left( 1-\ln x \right)}{\left( {{\ln }^{2}}x-{{x}^{2}} \right)\left( {{\ln }^{2}}x+{{x}^{2}} \right)} \right)}dx\].
\[\Rightarrow I=\dfrac{1}{2}\int{\left( \dfrac{\left( 1-\ln x \right)}{\left( {{\ln }^{2}}x-{{x}^{2}} \right)} \right)-\left( \dfrac{\left( 1-\ln x \right)}{\left( {{\ln }^{2}}x+{{x}^{2}} \right)} \right)}dx\] ----(2).
We know that $\int{f\left( x \right)+g\left( x \right)}dx=\int{f\left( x \right)dx}+\int{g\left( x \right)dx}$. We use this result in equation (2).
\[\Rightarrow I=\dfrac{1}{2}\left( \int{\dfrac{\left( 1-\ln x \right)}{\left( {{\ln }^{2}}x-{{x}^{2}} \right)}dx} \right)-\dfrac{1}{2}\left( \int{\dfrac{\left( 1-\ln x \right)}{\left( {{\ln }^{2}}x+{{x}^{2}} \right)}dx} \right)\].
Let us assume \[\int{\dfrac{\left( 1-\ln x \right)}{\left( {{\ln }^{2}}x-{{x}^{2}} \right)}dx}\] as ${{I}_{1}}$ and \[\int{\dfrac{\left( 1-\ln x \right)}{\left( {{\ln }^{2}}x+{{x}^{2}} \right)}dx}\] as ${{I}_{2}}$.
So, we get \[I=\dfrac{1}{2}{{I}_{1}}-\dfrac{1}{2}{{I}_{2}}\] ---(3).
Let us solve ${{I}_{1}}$ and ${{I}_{2}}$ separately and then substitute in equation (3).
Let us solve ${{I}_{1}}$ first. So, we have \[{{I}_{1}}=\int{\dfrac{\left( 1-\ln x \right)}{\left( {{\ln }^{2}}x-{{x}^{2}} \right)}dx}\].
Let us take ${{x}^{2}}$ common in denominator of the integrand.
\[\Rightarrow {{I}_{1}}=\int{\dfrac{\left( 1-\ln x \right)}{{{x}^{2}}\left( \dfrac{{{\ln }^{2}}x}{{{x}^{2}}}-1 \right)}dx}\].
\[\Rightarrow {{I}_{1}}=\int{\dfrac{\left( 1-\ln x \right)}{{{x}^{2}}\left( {{\left( \dfrac{\ln x}{x} \right)}^{2}}-1 \right)}dx}\] ---(4).
Let us assume $\dfrac{\ln x}{x}=t$ ---(5). We differentiate in both sides.
$\Rightarrow d\left( \dfrac{\ln x}{x} \right)=dt$.
We know that $d\left( \dfrac{u}{v} \right)=\dfrac{vdu-udv}{{{v}^{2}}}$.
$\Rightarrow \dfrac{\left( x \right)d\left( \ln x \right)-\left( \ln x \right)d\left( x \right)}{{{x}^{2}}}=dt$.
We know that $d\left( \ln x \right)=\dfrac{1}{x}dx$.
$\Rightarrow \dfrac{\left( x \right)\left( \dfrac{1}{x} \right)dx-\left( \ln x \right)dx}{{{x}^{2}}}=dt$.
$\Rightarrow \dfrac{\left( 1 \right)dx-\left( \ln x \right)dx}{{{x}^{2}}}=dt$.
$\Rightarrow \dfrac{\left( 1-\ln x \right)}{{{x}^{2}}}dx=dt$ ---(6).
Let us substitute equations (5) and (6) in equation (4).
\[\Rightarrow {{I}_{1}}=\int{\dfrac{dt}{{{t}^{2}}-1}}\].
\[\Rightarrow {{I}_{1}}=\int{\dfrac{dt}{{{t}^{2}}-{{1}^{2}}}}\].
We know that $\int{\dfrac{dx}{{{x}^{2}}-{{a}^{2}}}}=\dfrac{1}{2a}\ln \left( \dfrac{x-a}{x+a} \right)+C$.
\[\Rightarrow {{I}_{1}}=\dfrac{1}{2\left( 1 \right)}\ln \left( \dfrac{t-1}{t+1} \right)+{{C}_{1}}\].
\[\Rightarrow {{I}_{1}}=\dfrac{1}{2}\ln \left( \dfrac{t-1}{t+1} \right)+{{C}_{1}}\].
From equation (5), we have $t=\dfrac{\ln x}{x}$.
\[\Rightarrow {{I}_{1}}=\dfrac{1}{2}\ln \left( \dfrac{\left( \dfrac{\ln x}{x} \right)-1}{\left( \dfrac{\ln x}{x} \right)+1} \right)+{{C}_{1}}\].
\[\Rightarrow {{I}_{1}}=\dfrac{1}{2}\ln \left( \dfrac{\left( \dfrac{\ln x-x}{x} \right)}{\left( \dfrac{\ln x+x}{x} \right)} \right)+{{C}_{1}}\].
\[\Rightarrow {{I}_{1}}=\dfrac{1}{2}\ln \left( \dfrac{\ln x-x}{\ln x+x} \right)+{{C}_{1}}\] ---(7).
Now, let us solve ${{I}_{2}}$ first. So, we have \[{{I}_{2}}=\int{\dfrac{\left( 1-\ln x \right)}{\left( {{\ln }^{2}}x+{{x}^{2}} \right)}dx}\].
Let us take ${{x}^{2}}$ common in denominator of the integrand.
\[\Rightarrow {{I}_{2}}=\int{\dfrac{\left( 1-\ln x \right)}{{{x}^{2}}\left( \dfrac{{{\ln }^{2}}x}{{{x}^{2}}}+1 \right)}dx}\].
\[\Rightarrow {{I}_{2}}=\int{\dfrac{\left( 1-\ln x \right)}{{{x}^{2}}\left( {{\left( \dfrac{\ln x}{x} \right)}^{2}}+1 \right)}dx}\]---(8).
Let us assume $\dfrac{\ln x}{x}=t$ ---(9). We differentiate in both sides.
$\Rightarrow d\left( \dfrac{\ln x}{x} \right)=dt$.
We know that $d\left( \dfrac{u}{v} \right)=\dfrac{vdu-udv}{{{v}^{2}}}$.
$\Rightarrow \dfrac{\left( x \right)d\left( \ln x \right)-\left( \ln x \right)d\left( x \right)}{{{x}^{2}}}=dt$.
We know that $d\left( \ln x \right)=\dfrac{1}{x}dx$.
$\Rightarrow \dfrac{\left( x \right)\left( \dfrac{1}{x} \right)dx-\left( \ln x \right)dx}{{{x}^{2}}}=dt$.
$\Rightarrow \dfrac{\left( 1 \right)dx-\left( \ln x \right)dx}{{{x}^{2}}}=dt$.
$\Rightarrow \dfrac{\left( 1-\ln x \right)}{{{x}^{2}}}dx=dt$ ---(10).
Let us substitute equations (9) and (10) in equation (8).
\[\Rightarrow {{I}_{2}}=\int{\dfrac{dt}{{{t}^{2}}+1}}\].
\[\Rightarrow {{I}_{2}}=\int{\dfrac{dt}{{{t}^{2}}+{{1}^{2}}}}\].
We know that $\int{\dfrac{dx}{{{x}^{2}}+{{a}^{2}}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)+C$.
\[\Rightarrow {{I}_{2}}=\dfrac{1}{1}{{\tan }^{-1}}\left( \dfrac{t}{1} \right)+{{C}_{2}}\].
\[\Rightarrow {{I}_{2}}={{\tan }^{-1}}\left( t \right)+{{C}_{2}}\].
From equation (5), we have $t=\dfrac{\ln x}{x}$.
\[\Rightarrow {{I}_{2}}={{\tan }^{-1}}\left( \dfrac{\ln x}{x} \right)+{{C}_{2}}\] ---(11).
Let us substitute equations (7) and (11) in equation (3).
$\Rightarrow I=\dfrac{1}{2}\left( \dfrac{1}{2}\ln \left( \dfrac{\ln \left( x \right)-x}{\ln \left( x \right)+x} \right)+{{C}_{1}} \right)-\dfrac{1}{2}\left( {{\tan }^{-1}}\left( \dfrac{\ln x}{x} \right)+{{C}_{2}} \right)$.
$\Rightarrow I=\dfrac{1}{4}\ln \left( \dfrac{\ln \left( x \right)-x}{\ln \left( x \right)+x} \right)+\dfrac{{{C}_{1}}}{2}-\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{\ln x}{x} \right)-\dfrac{{{C}_{2}}}{2}$.
Let us $\dfrac{{{C}_{1}}}{2}-\dfrac{{{C}_{2}}}{2}$ as C, which is a constant of integration.
$\Rightarrow I=\dfrac{1}{4}\ln \left( \dfrac{\ln \left( x \right)-x}{\ln \left( x \right)+x} \right)-\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{\ln x}{x} \right)+C$.
So, we have found \[\int{\dfrac{{{x}^{2}}\left( 1-\ln x \right)}{{{\ln }^{4}}x-{{x}^{4}}}}dx=\dfrac{1}{4}\ln \left( \dfrac{\ln \left( x \right)-x}{\ln \left( x \right)+x} \right)-\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{\ln x}{x} \right)+C\].
The correct option for the given problem is (b).

Note: We should confuse ${{\ln }^{2}}x$ with $\ln {{x}^{2}}$, we get ${{\ln }^{2}}x$ by squaring the $\ln x$ i.e., ${{\ln }^{2}}x={{\left( \ln x \right)}^{2}}$. We should not confuse the formulas of integration. We have high chance of confusing the formula $\int{\dfrac{dx}{{{x}^{2}}-{{a}^{2}}}}=\dfrac{1}{2a}\ln \left( \dfrac{x-a}{x+a} \right)+C$ with $\int{\dfrac{dx}{{{a}^{2}}-{{x}^{2}}}}=\dfrac{1}{2a}\ln \left( \dfrac{a+x}{a-x} \right)+C$. We should not forget to re-substitute the function of t again in the integral again. Similarly, we can also expect problems to find the given integral with limits.