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Question: The integral \(\int {\left[ {\dfrac{{2{x^{12}} + 5{x^9}}}{{{{\left( {{x^5} + {x^3} + 1} \right)}^3}}...

The integral ∫[2x12+5x9(x5+x3+1)3]dx\int {\left[ {\dfrac{{2{x^{12}} + 5{x^9}}}{{{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} \right]dx} is equal to

A. x102(x5+x3+1)2+C B. x52(x5+x3+1)2+C C. −x102(x5+x3+1)2+C D. −x52(x5+x3+1)2+C  {\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 \\\
Explanation

Solution

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

Let the given integral be

I=∫[2x12+5x9(x5+x3+1)3]dx=∫[2x12+5x9[x5(x5+x3+1x5)]3]dx=∫[2x12+5x9[x5(x5x5+x3x5+1x5)]3]dx=∫[2x12+5x9[x5(1+x−2+x−5)]3]dx ⇒I=∫[2x12+5x9x15(1+x−2+x−5)3]dx=∫[2x12+5x9x15(1+x−2+x−5)3]dx  {\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 x15{x^{15}} common from the numerator also.
⇒I=∫[x15(2x12+5x9x15)x15(1+x−2+x−5)3]dx=∫[x15(2x12x15+5x9x15)x15(1+x−2+x−5)3]dx=∫[x15(2x−3+5x−6)x15(1+x−2+x−5)3]dx\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 x15{x^{15}} from the numerator with the x15{x^{15}} in the denominator, we get
⇒I=∫[2x−3+5x−6(1+x−2+x−5)3]dx →(1)\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=(1+x−2+x−5) →(2)t = \left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right){\text{ }} \to {\text{(2)}}
Let us differentiate equation (1) with respect to xx both sides, we get

dtdx=d(1+x−2+x−5)dx=0+d(x−2)dx+d(x−5)dx=−2x−3−5x−6=−(2x−3+5x−6) ⇒−dt=(2x−3+5x−6)dx →(3)  \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
⇒I=∫[−1t3]dt=−∫[t−3]dt=−[t−2−2]+C=12t2+C\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 CC is a constant of integration.
Now substitute the value of tt back in terms of xx using equation (2), we get
⇒I=12t2+C=12(1+x−2+x−5)2+C=12(1+1x2+1x5)2+C=12(x5+x3+1x5)2+C=x102(x5+x3+1)2+C\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, I=∫[2x12+5x9(x5+x3+1)3]dx=x102(x5+x3+1)2+C{\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.