Question

The integral $\int {\left[ {\dfrac{{2{x^{12}} + 5{x^9}}}{{{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} \right]dx} $ is equal to
\[
  {\text{A}}{\text{. }}\dfrac{{{x^{10}}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C \\
  {\text{B}}{\text{. }}\dfrac{{{x^5}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C \\
  {\text{C}}{\text{. }}\dfrac{{ - {x^{10}}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C \\
  {\text{D}}{\text{. }}\dfrac{{ - {x^5}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C \\
 \]

Answer 1

Hint- Here, we will be using integration by substitution method.

Let the given integral be \[
  {\text{I}} = \int {\left[ {\dfrac{{2{x^{12}} + 5{x^9}}}{{{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} \right]dx} = \int {\left[ {\dfrac{{2{x^{12}} + 5{x^9}}}{{{{\left[ {{x^5}\left( {\dfrac{{{x^5} + {x^3} + 1}}{{{x^5}}}} \right)} \right]}^3}}}} \right]dx} = \int {\left[ {\dfrac{{2{x^{12}} + 5{x^9}}}{{{{\left[ {{x^5}\left( {\dfrac{{{x^5}}}{{{x^5}}} + \dfrac{{{x^3}}}{{{x^5}}} + \dfrac{1}{{{x^5}}}} \right)} \right]}^3}}}} \right]dx} = \int {\left[ {\dfrac{{2{x^{12}} + 5{x^9}}}{{{{\left[ {{x^5}\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)} \right]}^3}}}} \right]dx} \\
   \Rightarrow {\text{I}} = \int {\left[ {\dfrac{{2{x^{12}} + 5{x^9}}}{{{x^{15}}{{\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}^3}}}} \right]dx} = \int {\left[ {\dfrac{{2{x^{12}} + 5{x^9}}}{{{x^{15}}{{\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}^3}}}} \right]dx} \\
 \]
In the above integral, let us take ${x^{15}}$ common from the numerator also.
\[ \Rightarrow {\text{I}} = \int {\left[ {\dfrac{{{x^{15}}\left( {\dfrac{{2{x^{12}} + 5{x^9}}}{{{x^{15}}}}} \right)}}{{{x^{15}}{{\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}^3}}}} \right]dx} = \int {\left[ {\dfrac{{{x^{15}}\left( {\dfrac{{2{x^{12}}}}{{{x^{15}}}} + \dfrac{{5{x^9}}}{{{x^{15}}}}} \right)}}{{{x^{15}}{{\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}^3}}}} \right]dx} = \int {\left[ {\dfrac{{{x^{15}}\left( {2{x^{ - 3}} + 5{x^{ - 6}}} \right)}}{{{x^{15}}{{\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}^3}}}} \right]dx} \]
Now let us cancel out ${x^{15}}$ from the numerator with the ${x^{15}}$ in the denominator, we get
\[ \Rightarrow {\text{I}} = \int {\left[ {\dfrac{{2{x^{ - 3}} + 5{x^{ - 6}}}}{{{{\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}^3}}}} \right]dx} {\text{ }} \to {\text{(1)}}\]
In order to solve the above integral, we will use integration by substitution method.
Put \[t = \left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right){\text{ }} \to {\text{(2)}}\]
Let us differentiate equation (1) with respect to $x$ both sides, we get
\[
  \dfrac{{dt}}{{dx}} = \dfrac{{d\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}}{{dx}} = 0 + \dfrac{{d\left( {{x^{ - 2}}} \right)}}{{dx}} + \dfrac{{d\left( {{x^{ - 5}}} \right)}}{{dx}} = - 2{x^{ - 3}} - 5{x^{ - 6}} = - \left( {2{x^{ - 3}} + 5{x^{ - 6}}} \right) \\
   \Rightarrow - dt = \left( {2{x^{ - 3}} + 5{x^{ - 6}}} \right)dx{\text{ }} \to {\text{(3)}} \\
 \]
Clearly, we can see that after differentiating the assumed function we are getting the numerator of the integral that we are supposed to find.
Using equation (2) and (3) in equation (1), the integral becomes
\[ \Rightarrow {\text{I}} = \int {\left[ {\dfrac{{ - 1}}{{{t^3}}}} \right]dt} = - \int {\left[ {{t^{ - 3}}} \right]dt} = - \left[ {\dfrac{{{t^{ - 2}}}}{{ - 2}}} \right] + C = \dfrac{1}{{2{t^2}}} + C\] where $C$ is a constant of integration.
Now substitute the value of $t$ back in terms of $x$ using equation (2), we get
\[ \Rightarrow {\text{I}} = \dfrac{1}{{2{t^2}}} + C = \dfrac{1}{{2{{\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}^2}}} + C = \dfrac{1}{{2{{\left( {1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^5}}}} \right)}^2}}} + C = \dfrac{1}{{2{{\left( {\dfrac{{{x^5} + {x^3} + 1}}{{{x^5}}}} \right)}^2}}} + C = \dfrac{{{x^{10}}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C\]
Therefore, ${\text{I}} = \int {\left[ {\dfrac{{2{x^{12}} + 5{x^9}}}{{{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} \right]dx} = \dfrac{{{x^{10}}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C$
Hence, option A is correct.

Note- In this problem, we have finally converted the integral in a form where the differentiation of the denominator function gives the numerator function and then by putting the denominator function as another variable, the given integral is solved.