Question

Integrate the following equation: \[\int {\dfrac{{2{x^3} - 3{x^2} - 8x - 26}}{{2{x^2} - 5x + 2}}dx} \].
A. \[\dfrac{{{x^2}}}{2} + x - \dfrac{{38}}{3}\log \left| {x - 2} \right| + \dfrac{{61}}{6}\log \left| {2x - 1} \right| + C\]
B. Doesn’t exist
C. Cannot be determined
D. None of these

Answer 1

Hint: The given problem revolves around the concepts of integration. Keeping in mind, first of all dividing both the equations i.e. \[2{x^3} - 3{x^2} - 8x - 26\] and \[2{x^2} - 5x + 2\] respectively, as the dividend and divisor. Then, by separating the certain terms and using the partial fraction method i.e. \[\dfrac{{px + q}}{{\left( {x + a} \right)\left( {x + b} \right)\left( {x + c} \right)}} = \dfrac{A}{{x + a}} + \dfrac{B}{{x + b}} + \dfrac{C}{{x + c}}\]. As a result, by using the formulae (or, rules) of the integration for the n terms such as \[\int {\dfrac{1}{x}dx} = \log x + c\], etc. and then substituting the values in the given equation, the desire solution is obtained.

Complete step by step answer:
Since, we have given the expression that
\[ \Rightarrow \int {\dfrac{{2{x^3} - 3{x^2} - 8x - 26}}{{2{x^2} - 5x + 2}}dx} \] … (i)
Since, the power/degree is greater in numerator than that of the denominator,
Hence, simplifying the equation mathematically that is dividing the equation, we get,
\[2{x}^2-5x+2\overset{x+1}{\overline{\left){\begin{align}
  & {2{x}^{3}}-3{{x}^{2}}-8x-26 \\
 & \underline{{2{x}^{3}}-5{{x}^{2}+2x}} \\
 & 2{{x}^{2}}-10x-26 \\
 & \underline{2{{x}^{2}}-5x+2} \\
 & -5x-28 \\
 & \underline{6x-24} \\
 & 0 \\
\end{align}}\right.}}\]
Quotient equals to \[x + 1\],
Remainder equals to \[ - 5x - 28\],
Dividend equals to \[2{x^3} - 3{x^2} - 8x - 26\],
And,
Divisor equals \[2{x^2} - 5x + 2\].
Hence, the equation (i) can be written/equals to
\[ \Rightarrow \int {\dfrac{{2{x^3} - 3{x^2} - 8x - 26}}{{2{x^2} - 5x + 2}}dx} = \int {\left( {\left( {x + 1} \right) - \dfrac{{\left( {5x + 28} \right)}}{{2{x^2} - 5x + 2}}} \right)} dx\]
… (\[\because \dfrac{{{\text{Dividend}}}}{{{\text{Divisor}}}}{\text{ = Quotient}} \pm \dfrac{{{\text{Remainder}}}}{{{\text{Divisor}}}}\])
Separating the terms so as to integrate the equation, we get
\[ \Rightarrow \int {\dfrac{{2{x^3} - 3{x^2} - 8x - 26}}{{2{x^2} - 5x + 2}}dx} = \int {xdx} + \int {dx} - \int {\dfrac{{5x + 28}}{{2{x^2} - 5x + 2}}} dx\]
\[ \Rightarrow \int {\dfrac{{2{x^3} - 3{x^2} - 8x - 26}}{{2{x^2} - 5x + 2}}dx} = \int {xdx} + \int {dx} - \int {\dfrac{{5x + 28}}{{\left( {x - 2} \right)\left( {2x - 1} \right)}}} dx\]
… (\[\because 2{x^2} - 5x + 2 = 2{x^2} - 4x - x + 2 = 2x\left( {x - 2} \right) - 1\left( {x - 2} \right) = \left( {x - 2} \right)\left( {2x - 1} \right)\])
Hence, using the formula \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c} \] to find out the integration of first two terms in the above equation, we get
\[ \Rightarrow \int {\dfrac{{2{x^3} - 3{x^2} - 8x - 26}}{{2{x^2} - 5x + 2}}dx} = \dfrac{{{x^2}}}{2} + x - \int {\dfrac{{5x + 28}}{{\left( {x - 2} \right)\left( {2x - 1} \right)}}} dx + {c_1} + {c_2}\] … (ii)
Where;
\[{c_1},{c_2}\] are the constants of first two (solved) integration,
Since, for \[\int {\dfrac{{5x + 28}}{{\left( {x - 2} \right)\left( {2x - 1} \right)}}} dx\] in the above equation,
$\because $We know that,
Integration for $\int {\dfrac{{px + q}}{{\left( {x - a} \right)\left( {x - b} \right)}}dx = } \int {\left( {\dfrac{A}{{x - a}} + \dfrac{B}{{\left( {x - b} \right)}}} \right)dx} $, we get
As a result, equating the certain terms (inside the integration sign), we get
\[\therefore \dfrac{{5x + 28}}{{\left( {x - 2} \right)\left( {2x - 1} \right)}} = \dfrac{A}{{x - 2}} + \dfrac{B}{{2x - 1}}\] … (iii)
\[\dfrac{{5x + 28}}{{\left( {x - 2} \right)\left( {2x - 1} \right)}} = \dfrac{{A\left( {2x - 1} \right) + B\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {2x - 1} \right)}}\]
Hence, the equation becomes
\[5x + 28 = A\left( {2x - 1} \right) + B\left( {x - 2} \right)\]
\[5x + 28 = 2Ax - A + Bx - 2B\]
Hence, equating the respective terms (or, equation), we get
\[5x + 28 = \left( {2A + B} \right)x + \left( { - A - 2B} \right)\]
Comparing the equation i.e. in quadratic form, we get
\[2A + B = 5\] … (iv)
And,
\[ - A - 2B = 28\] … (v)
As a result, solving these equations (iv) and (v) for the desire value, we get
Therefore, since, multiplying the equation (v) by \[2\], we get
\[ - 2A - 4B = 56\] … (vi)
Hence, adding the equations (iv) and (vi) simultaneously, we get
\[ - 3B = 61\]
\[B = - \dfrac{{61}}{3}\]
Now, hence substituting \[B = - \dfrac{{61}}{3}\] in equation (iv), we get
\[2A - \dfrac{{61}}{3} = 5\]
\[2A = 5 + \dfrac{{61}}{3} = \dfrac{{76}}{3}\]
The value of ‘A’ is,
\[A = \dfrac{{38}}{3}\]
Now,
Hence, substituting the values i.e. ‘\[A = \dfrac{{38}}{3}\]’ and ‘\[B = - \dfrac{{61}}{3}\]’ equation (iii), we get
\[\therefore \dfrac{{5x + 28}}{{\left( {x - 2} \right)\left( {2x - 1} \right)}} = \dfrac{{\dfrac{{38}}{3}}}{{x - 2}} - \dfrac{{\dfrac{{61}}{3}}}{{2x - 1}}\]
Now, hence substituting these values i.e. \[\dfrac{{5x + 28}}{{\left( {x - 2} \right)\left( {2x - 1} \right)}}\] in the equation (ii), we get
\[ \Rightarrow \int {\dfrac{{2{x^3} - 3{x^2} - 8x - 26}}{{2{x^2} - 5x + 2}}dx} = \dfrac{{{x^2}}}{2} + x - \int {\left( {\dfrac{{\dfrac{{38}}{3}}}{{x - 2}} - \dfrac{{\dfrac{{61}}{3}}}{{2x - 1}}} \right)} dx + {c_1} + {c_2}\]
Again, separating the terms in the integration (as done previously), we get
\[ \Rightarrow \int {\dfrac{{2{x^3} - 3{x^2} - 8x - 26}}{{2{x^2} - 5x + 2}}dx} = \dfrac{{{x^2}}}{2} + x - \dfrac{{38}}{3}\int {\dfrac{1}{{x - 2}}dx} + \dfrac{{61}}{3}\int {\dfrac{1}{{2x - 1}}} dx + {c_1} + {c_2}\]
Since, for the term \[\dfrac{1}{{2x - 1}}\], multiply and divide by ‘\[2\]’ in the above equation, so as to get the exact derivative of denominator in the numerator, we get
\[ \Rightarrow \int {\dfrac{{2{x^3} - 3{x^2} - 8x - 26}}{{2{x^2} - 5x + 2}}dx} = \dfrac{{{x^2}}}{2} + x - \dfrac{{38}}{3}\int {\dfrac{1}{{x - 2}}dx} + \dfrac{{61}}{6}\int {\dfrac{2}{{2x - 1}}} dx + {c_1} + {c_2}\]
Now,
Here, 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
Hence, implicating the formula, we get
\[ \Rightarrow \int {\dfrac{{2{x^3} - 3{x^2} - 8x - 26}}{{2{x^2} - 5x + 2}}dx} = \dfrac{{{x^2}}}{2} + x - \dfrac{{38}}{3}\log \left| {x - 2} \right| + \dfrac{{61}}{6}\log \left| {2x - 1} \right| + {c_1} + {c_2} + {c_3} + {c_4}\]
\[ \Rightarrow \int {\dfrac{{2{x^3} - 3{x^2} - 8x - 26}}{{2{x^2} - 5x + 2}}dx} = \dfrac{{{x^2}}}{2} + x - \dfrac{{38}}{3}\log \left| {x - 2} \right| + \dfrac{{61}}{6}\log \left| {2x - 1} \right| + C\]
Where, ${c_1},{c_2},{c_3},{c_4},...,{c_n}$ are the integration constants for the respective equation.
Hence, the solution!
$\therefore \Rightarrow $Option A is correct.6

Note: Particularly; in this case, remember that when the highest power in the numerator is greater than that of denominator, then divide the respective equations as we used to solve in previous classes i.e. by dividing the terms considering all the parameters such as quotient, remainder, dividend, divisor, etc. As a result, use of the formulae/condition that is “Integration by using Partial Fraction method” i.e. \[\dfrac{{px + q}}{{\left( {x + a} \right)\left( {x + b} \right)}} = \dfrac{A}{{x + a}} + \dfrac{B}{{x + b}}\] respectively (i.e. used when the derivatives contains the non-repeated linear factors). Also, remember the formulae such as $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$, \[\int {\dfrac{1}{x}dx} = \log x + c\], \[\int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}dx} = \log \left[ {f'\left( x \right)} \right] + c\], etc., so as to be sure of our final answer.