Question

Integrate the following equation: $\int {\dfrac{1}{{{x^3}\left( {{x^3} - 1} \right)}}dx} $.
(a) $\dfrac{1}{3}\log \left( {x - 1} \right) - \dfrac{1}{6}\log \left( {{x^2} + x + 1} \right) - \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{2x + 1}}{{\sqrt 3 }}} \right) + c$
(b) $\dfrac{1}{3}\log \left( {x - 1} \right) - \dfrac{1}{6}\log \left( {{x^2} + x + 1} \right) - \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{2x + 1}}{{\sqrt 3 }}} \right) + \dfrac{1}{{2{x^2}}} + c$
(c) $ - \dfrac{1}{3}\log \left( {x - 1} \right) + \dfrac{1}{6}\log \left( {{x^2} + x + 1} \right) - \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{2x + 1}}{{\sqrt 3 }}} \right) - \dfrac{1}{{2{x^2}}} + c$
(d) None of these

Answer 1

Hint: The given problem revolves around the concepts of integration. Keeping in mind, first of all making the numerator as where we can separate the denominator, say, ${x^3} - \left( {{x^3} - 1} \right) = 1$. Then, the first term will be in complex condition which can be solving again by separating the numerator and denominator respectively, (will be in $\int {\dfrac{{dx}}{{\left( {x - a} \right)\left( {{x^2} + x + b} \right)}} = } \int {\left( {\dfrac{A}{{x - a}} + \dfrac{{Bx + C}}{{\left( {{x^2} + x + b} \right)}}} \right)dx} $ mode). Hence, finding the constant(s) A,B and C puts it in the condition. As a result, obtained the remaining terms by using the formulae (or, rules) of the integration for the n terms such as \[\int {\dfrac{1}{x}dx} = \log x + c\], and then substituting the values in the given expression, to obtained the desire solution.

Complete step-by-step answer:
Since, we have given the expression that
$\int {\dfrac{1}{{{x^3}\left( {{x^3} - 1} \right)}}dx} $
Simplifying the expression mathematically that is ${x^3} - \left( {{x^3} - 1} \right) = 1$, we get
$\int {\dfrac{1}{{{x^3}\left( {{x^3} - 1} \right)}}dx} = \int {\dfrac{{{x^3} - \left( {{x^3} - 1} \right)}}{{{x^3}\left( {{x^3} - 1} \right)}}dx} $
Separating the numerator and denominator, we get
$\int {\dfrac{1}{{{x^3}\left( {{x^3} - 1} \right)}}dx} = \int {\dfrac{{{x^3}}}{{{x^3}\left( {{x^3} - 1} \right)}}dx} - \int {\dfrac{{{x^3} - 1}}{{{x^3}\left( {{x^3} - 1} \right)}}dx} $
$\int {\dfrac{1}{{{x^3}\left( {{x^3} - 1} \right)}}dx} = \int {\dfrac{1}{{{x^3} - 1}}dx} - \int {\dfrac{1}{{{x^3}}}dx} $ … (i)
Now, let us find the integration by separating both the terms,
Therefore, let
${I_1} = \int {\dfrac{{dx}}{{{x^3} - 1}}} $
And,
${I_2} = \int {\dfrac{{dx}}{{{x^3}}}} $$\int {\dfrac{{dx}}{{\left( {x - a} \right)\left( {{x^2} + x + b} \right)}} = } \int {\left( {\dfrac{A}{{x - a}} + \dfrac{{Bx + C}}{{\left( {{x^2} + x + b} \right)}}} \right)dx} $
Equation (i) becomes
$\int {\dfrac{1}{{{x^3}\left( {{x^3} - 1} \right)}}dx} = {I_1} - {I_2}$ … (ii)
First of solving ‘${I_1}$’, we get
${I_1} = \int {\dfrac{{dx}}{{{x^3} - 1}}} = \int {\dfrac{{dx}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}} $
$\because $We know that,
Integration for
The equation becomes,
\[{I_1} = \int {\dfrac{1}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}dx} = \int {\left( {\dfrac{A}{{x - 1}} + \dfrac{{Bx + C}}{{{x^2} + x + 1}}} \right)dx} \] … (iii)
Solving the equation mathematically, we get
\[\dfrac{1}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{A}{{x - 1}} + \dfrac{{Bx + C}}{{{x^2} + x + 1}} = \dfrac{{A\left( {{x^2} + x + 1} \right) + \left( {Bx + C} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\]
Hence, equating the terms (or, equation), we get
\[1 = A\left( {{x^2} + x + 1} \right) + \left( {Bx + C} \right)\left( {x - 1} \right)\]
\[1 = A{x^2} + Ax + A + B{x^2} - Bx + Cx - C\]
Comparing the equation i.e. in quadratic form, we get
\[\left( 0 \right){x^2} + \left( 0 \right)x + 1 = \left( {A + B} \right){x^2} + \left( {A - B + C} \right)x + \left( {A - C} \right)\]
Equating the respective terms of certain quadratic equation i.e. of \[{x^2}\], $x$ and ${\text{constant}}$ respectively, we get
\[A + B = 0\] … (iv)
$A - B + C = 0$ … (v)
And, $A - C = 1$ … (vi)
As a result, solving these equations (iv), (v) and (vi) for the desire value, we get
Consider, equation (iv)
\[A + B = 0\]
$B = - A$ … (vii)
Hence, substituting $B = - A$ in equation (v), we get
$A + A + C = 0$
$2A + C = 0$ … (viii)
Now,
$\therefore $Solving i.e. adding the equations (vi) and (viii), we get
$3A = 1$
$A = \dfrac{1}{3}$
Substitute $A = \dfrac{1}{3}$ in equation (vii) and (vi) respectively, we get
$B = - \dfrac{1}{3}$
And,
\[\dfrac{1}{3} - 1 = C\]
\[C = \dfrac{{ - 2}}{3}\]
Now,
Hence, substituting all the values i.e. ‘A’, ‘B’ and ‘C’ equation (ii), we get
\[{I_1} = \int {\dfrac{1}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}dx} = \int {\left( {\dfrac{{\dfrac{1}{3}}}{{x - 1}} + \dfrac{{ - \dfrac{1}{3}x - \dfrac{2}{3}}}{{{x^2} + x + 1}}} \right)dx} \]
Solving the equation mathematically, we get
\[{I_1} = \int {\left[ {\dfrac{1}{{3\left( {x - 1} \right)}} - \dfrac{1}{3}\left( {\dfrac{{x + 2}}{{{x^2} + x + 1}}} \right)} \right]dx} \]
Separating the terms in the integration, we get
\[{I_1} = \dfrac{1}{3}\int {\dfrac{{dx}}{{x - 1}}} - \dfrac{1}{3}\int {\left( {\dfrac{{x + 2}}{{{x^2} + x + 1}}} \right)dx} \]
For separating the integration (to solve the equation easily),
\[{I_1} = {I_A} - {I_B}\] … (ix)
Where, ${I_A} = \dfrac{1}{3}\int {\dfrac{{dx}}{{x - 1}}} $ … (x)
And,
${I_B} = \dfrac{1}{3}\int {\left( {\dfrac{{x + 2}}{{{x^2} + x + 1}}} \right)dx} $ … (xi)
Considering the equation (x),
${I_A} = \dfrac{1}{3}\int {\dfrac{{dx}}{{x - 1}}} $
Since, we know that
\[\int {\dfrac{1}{x}dx} = \log x + c\]
\[\therefore \int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}dx} = \log \left[ {f'\left( x \right)} \right] + c\], we get
The equation (x) becomes,
${I_A} = \dfrac{1}{3}\log \left( {x - 1} \right) + {c_1}$ … (xii)
Where, ${c_1},{c_2},{c_3},....{c_n}$ are integration constants respectively!
Now, considering the equation (xi),
${I_B} = \dfrac{1}{3}\int {\left( {\dfrac{{x + 2}}{{{x^2} + x + 1}}} \right)dx} $
Multiply and divide the equation by $2$, we get
${I_B} = \dfrac{1}{6}\int {\left( {\dfrac{{2x + 4}}{{{x^2} + x + 1}}} \right)dx} $
Term inside the integration i.e. $2x + 4$ can also be written as,
${I_B} = \dfrac{1}{6}\int {\left( {\dfrac{{\left( {2x + 1} \right) + 3}}{{{x^2} + x + 1}}} \right)dx} $
Separating the terms, we get
${I_B} = \dfrac{1}{6}\int {\left( {\dfrac{{2x + 1}}{{{x^2} + x + 1}}} \right)dx} + \dfrac{3}{6}\int {\dfrac{1}{{{x^2} + x + 1}}dx} $
Now, hence it seems that in the first term, numerator is the exact derivative of the denominator i.e.
$f'\left( {{x^2} + x + 1} \right) = 2x + 1$
\[\therefore \int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}dx} = \log \left[ {f'\left( x \right)} \right] + c\], we get
\[{I_B} = \dfrac{1}{6}\log \left( {{x^2} + x + 1} \right) + {c_2} + \dfrac{1}{2}\int {\dfrac{1}{{{x^2} + x + 1}}dx} \]
And, since considering the second term i.e. the denominator ${x^2} + x + 1$ we can also write ${\left( {x + \dfrac{1}{2}} \right)^2} + {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2}$ which seems that \[{\left( {x + \dfrac{1}{2}} \right)^2} + {\left( {\dfrac{{\sqrt 3 }}{4}} \right)^2} = {x^2} + 2x\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4} = {x^2} + x + 1\], we get
Hence, the equation (xi) becomes
\[{I_B} = \dfrac{1}{6}\log \left( {{x^2} + x + 1} \right) + {c_1}\dfrac{1}{2}\int {\dfrac{1}{{{{\left( {x + \dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}}}dx} \]
Now, we know that
$\int {\dfrac{1}{{{x^2} + {a^2}}}dx} = \dfrac{1}{a}{\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right) + c$
Hence, comparatively the equation becomes
\[{I_B} = \dfrac{1}{6}\log \left( {{x^2} + x + 1} \right) + {c_2} + \dfrac{1}{2} \times \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}}{\tan ^{ - 1}}\left( {\dfrac{{x + \dfrac{1}{2}}}{{\dfrac{{\sqrt 3 }}{2}}}} \right) + {c_3}\]
\[{I_B} = \dfrac{1}{6}\log \left( {{x^2} + x + 1} \right) + {c_2} + \dfrac{1}{2} \times \dfrac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{2x + 1}}{2}}}{{\dfrac{{\sqrt 3 }}{2}}}} \right) + {c_3}\]
As a result, solving the equation mathematically, we get
\[{I_B} = \dfrac{1}{6}\log \left( {{x^2} + x + 1} \right) + {c_1} + \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{2x + 1}}{{\sqrt 3 }}} \right) + {c_2}\] … (xiii)
From (xii) and (xiii),
Equation (ix) becomes,
\[{I_1} = \dfrac{1}{3}\log \left( {x - 1} \right) + {c_1} - \dfrac{1}{6}\log \left( {{x^2} + x + 1} \right) + {c_2} - \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{2x + 1}}{{\sqrt 3 }}} \right) + {c_3}\] … (xiv)
Similarly,
Solving ‘${I_2}$’ i.e. ${I_2} = \int {\dfrac{{dx}}{{{x^3}}}} $, we get
${I_2} = \int {\dfrac{{dx}}{{{x^3}}}} = \int {{x^{ - 3}}dx} $
Since, we know that
\[\int {{x^n}dx} = \dfrac{{{x^n}}}{{{x^{n + 1}}}} + c\] where, ${c_1},{c_2},{c_3},....{c_n}$ are integration constants respectively!
Hence, the equation becomes
\[{I_2} = \int {\dfrac{{{x^{ - 3 + 1}}}}{{ - 3 + 1}}dx} \]
\[{I_2} = \int {\dfrac{{{x^{ - 2}}}}{{ - 2}}dx} \]
Mathematically solving the equation, we get
\[{I_2} = - \dfrac{1}{2}\int {\dfrac{1}{{{x^2}}}dx} \]
\[{I_2} = - \dfrac{1}{{2{x^2}}} + {c_4}\] … (xv)
As a result, from (xiv) and (xv)
Equation (i) becomes, we get
$\int {\dfrac{1}{{{x^3}\left( {{x^3} - 1} \right)}}dx} = \dfrac{1}{3}\log \left( {x - 1} \right) + {c_1} - \dfrac{1}{6}\log \left( {{x^2} + x + 1} \right) + {c_2} - \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{2x + 1}}{{\sqrt 3 }}} \right) + {c_3} - \left( { - \dfrac{1}{{2{x^2}}} + {c_4}} \right)$$\int {\dfrac{1}{{{x^3}\left( {{x^3} - 1} \right)}}dx} = \dfrac{1}{3}\log \left( {x - 1} \right) + {c_1} - \dfrac{1}{6}\log \left( {{x^2} + x + 1} \right) + {c_2} - \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{2x + 1}}{{\sqrt 3 }}} \right) + {c_3} + \dfrac{1}{{2{x^2}}} - {c_4}$
Since, assuming $c = {c_1},{c_2},{c_3},.....$ as a constants of all the integration, we get
$\int {\dfrac{1}{{{x^3}\left( {{x^3} - 1} \right)}}dx} = \dfrac{1}{3}\log \left( {x - 1} \right) - \dfrac{1}{6}\log \left( {{x^2} + x + 1} \right) - \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{2x + 1}}{{\sqrt 3 }}} \right) + \dfrac{1}{{2{x^2}}} + c$
$\therefore \Rightarrow $Option (b) is correct.
So, the correct answer is “Option b”.

Note: One must remember the concept of integration and its simultaneous formulae. As a result, to get the desired outcome, remember the condition i.e. $\int {\dfrac{{dx}}{{\left( {x - a} \right)\left( {{x^2} + x + b} \right)}} = } \int {\left( {\dfrac{A}{{x - a}} + \dfrac{{Bx + C}}{{\left( {{x^2} + x + b} \right)}}} \right)dx} $ in this case particular. Algebraically solving the equations/solution use (or, substitute) the formulae like $\int {\dfrac{1}{{{x^2} + {a^2}}}dx} = \dfrac{1}{a}{\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right) + c$, \[\int {\dfrac{1}{x}dx} = \log x + c\], \[\int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}dx} = \log \left[ {f'\left( x \right)} \right] + c\], etc., so as to be sure of our final answer.