Question

If $\int {\dfrac{{dx}}{{{x^4} + {x^3}}} = \dfrac{A}{{{x^2}}} + \dfrac{B}{x} + \log \left| {\dfrac{x}{{x + 1}}} \right| + c} $, then,
A. $A = \dfrac{1}{2},B = 1$
B. $A = 1,B = \dfrac{1}{2}$
C. $A = - \dfrac{1}{2},B = 1$
D. $A = 1,B = 1$

Answer 1

Hint: To solve the integration, we will first take ${x^3}$ common from the denominator. Then by using partial fraction, we will write the term to be integrated in the forms of partial fractions. Then by integrating the partial fraction we will get an answer. To find the value of $A$ and $B$, we will compare the answer of integration we obtained, with the given answer.

Complete step by step answer:
Given, $\int {\dfrac{{dx}}{{{x^4} + {x^3}}} = \dfrac{A}{{{x^2}}} + \dfrac{B}{x} + \log \left| {\dfrac{x}{{x + 1}}} \right| + c} $.
LHS:- $\int {\dfrac{{dx}}{{{x^4} + {x^3}}}} $
Taking ${x^3}$ common in the denominator, we get,
\[ \int {\dfrac{{dx}}{{{x^3}\left( {x + 1} \right)}}} \]
Now, writing the term as partial fraction, we can write,
\[\dfrac{1}{{{x^3}\left( {x + 1} \right)}} = \dfrac{P}{x} + \dfrac{Q}{{{x^2}}} + \dfrac{R}{{{x^3}}} + \dfrac{S}{{\left( {x + 1} \right)}}\]
Taking, LCM of terms on right hand side, gives us,
$ \Rightarrow \dfrac{1}{{{x^3}\left( {x + 1} \right)}} = \dfrac{{{x^2}\left( {x + 1} \right)P + x\left( {x + 1} \right)Q + \left( {x + 1} \right)R + {x^3}S}}{{{x^3}\left( {x + 1} \right)}}$
Now, cancelling the denominator on both the sides, gives us,
$ \Rightarrow 1 = {x^2}\left( {x + 1} \right)P + x\left( {x + 1} \right)Q + \left( {x + 1} \right)R + {x^3}S - - - \left( 1 \right)$
Now, taking $x = - 1$, we get,
$1 = {\left( { - 1} \right)^2}\left( { - 1 + 1} \right)P + \left( { - 1} \right)\left( { - 1 + 1} \right)Q + \left( { - 1 + 1} \right)R + {\left( { - 1} \right)^3}S$
$ \Rightarrow 1 = 0 + 0 + 0 - S$
Multiplying both sides with $ - 1$, we get,
$ \Rightarrow S = - 1$
Now, taking $x = 0$, we get,
$1 = 0\left( {0 + 1} \right)P + 0\left( {0 + 1} \right)Q + \left( {0 + 1} \right)R + 0S$
$ \Rightarrow 1 = 0 + 0 + R + 0$
Now, changing the sides, we get,
$ \Rightarrow R = 1$
Now, comparing the coefficient of ${x^3}$ in both sides of $\left( 1 \right)$, we get,
$P + S = 0$
Subtracting both sides with $S$, we get,
$ \Rightarrow P = - S = - \left( { - 1} \right) = 1$
[Since, $S = - 1$, we already found above]
Now, comparing the coefficient of ${x^2}$ in both sides of $\left( 1 \right)$, we get,
$P + Q = 0$
Subtracting both sides with $P$, we get,
$ \Rightarrow Q = - P = - 1$
[Since, $P = 1$, we found above]
Therefore, $P = 1,Q = - 1,R = 1,S = - 1$.
So, we can write,
\[\dfrac{1}{{{x^3}\left( {x + 1} \right)}} = \dfrac{1}{x} - \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}} - \dfrac{1}{{\left( {x + 1} \right)}}\]
Therefore, integrating now,
$\int {\left[ {\dfrac{1}{x} - \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}} - \dfrac{1}{{\left( {x + 1} \right)}}} \right]} dx$
Now, opening the bracket,
$ \int {\dfrac{1}{x}dx - \int {\dfrac{1}{{{x^2}}}dx} + \int {\dfrac{1}{{{x^3}}}dx - \int {\dfrac{1}{{\left( {x + 1} \right)}}dx} } } $
Now, we know, $\int {\dfrac{1}{x}dx = \log x} $
And, $\int {{x^n}dx = } \dfrac{{{x^{n + 1}}}}{{n + 1}}$
Therefore, the following integration can be written as,
$\int {\dfrac{1}{x}dx - \int {{x^{ - 2}}dx} + \int {{x^{ - 3}}dx - \int {\dfrac{1}{{\left( {x + 1} \right)}}dx} } } $
$\Rightarrow \log \left| x \right| - \left[ {\dfrac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}}} \right] + \left[ {\dfrac{{{x^{ - 3 + 1}}}}{{ - 3 + 1}}} \right] - \int {\dfrac{1}{{(x + 1)}}dx} $
Here, for, $\int {\dfrac{1}{{x + 1}}dx} $,
Let $x + 1 = u$. Differentiating both sides, we get,
$dx = du$
Substituting these values, we get,
$\int {\dfrac{1}{u}du} = \log \left| u \right|$
$\Rightarrow \int {\dfrac{1}{u}du} = \log \left| {x + 1} \right|$
Therefore, the integration becomes,
$ \log \left| x \right| - \left[ {\dfrac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}}} \right] + \left[ {\dfrac{{{x^{ - 3 + 1}}}}{{ - 3 + 1}}} \right] - \log \left| {x + 1} \right| + c$
Where, $c = $constant of integration
$\log \left| x \right| - \dfrac{1}{{\left( { - 1} \right)x}} + \dfrac{1}{{\left( { - 2} \right){x^2}}} - \log \left| {x + 1} \right| + c$
Now, simplifying, we get,
$ \log \left| x \right| + \dfrac{1}{x} - \dfrac{1}{{2{x^2}}} - \log \left| {x + 1} \right| + c$
We know, $\log \left| a \right| - \log \left| b \right| = \log \left| {\dfrac{a}{b}} \right|$, using this property, we get,
$ \log \left| {\dfrac{x}{{x + 1}}} \right| + \dfrac{1}{x} - \dfrac{1}{{2{x^2}}} + c$
$\Rightarrow - \dfrac{1}{{2{x^2}}} + \dfrac{1}{x} + \log \left| {\dfrac{x}{{x + 1}}} \right| + c$
$\Rightarrow \int {\dfrac{{dx}}{{{x^4} + {x^3}}}} = - \dfrac{1}{{2{x^2}}} + \dfrac{1}{x} + \log \left| {\dfrac{x}{{x + 1}}} \right| + c$

RHS:-
$\dfrac{A}{{{x^2}}} + \dfrac{B}{x} + \log \left| {\dfrac{x}{{x + 1}}} \right| + c$
Comparing, LHS and RHS, we get,
$ - \dfrac{1}{{2{x^2}}} + \dfrac{1}{x} + \log \left| {\dfrac{x}{{x + 1}}} \right| + c = \dfrac{A}{{{x^2}}} + \dfrac{B}{x} + \log \left| {\dfrac{x}{{x + 1}}} \right| + c$
Comparing the coefficient of $\dfrac{1}{{{x^2}}}$, we get,
$A = - \dfrac{1}{2}$
And, comparing the coefficient of $\dfrac{1}{x}$, we get,
$B = 1$
Therefore, $A = - \dfrac{1}{2},B = 1$,

Hence the correct option is C.

Note: Sometimes integrations can feel really complex, like the one in this question, because we were not able to deduce this in the general form of integration. So, in such cases partial fraction is very useful, as using partial fraction we can deduce the term to be integrated into terms that are easy to integrate. Simplification rules must be followed to simplify the expressions.