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**Hint:**Start with trying to take common ${x^5}$from the denominator term$\left( {{x^5} + {x^3} + 1} \right)$. This will give you ${x^{15}}$in the denominator and then you can divide the numerator with${x^{15}}$. This will give you an expression$\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)$ on which you can integrate by using a substitution method.

**Complete step-by-step answer:**

According to the substitution rule in integration\[\int {f(g(x))g\prime (x)dx} = \int {f(u)du} ,where,u = g(x)\]

The substitution method is used when an integral contains some function and its derivative. In this case, we can set $m$ equal to the function and rewrite the integral in terms of the new variable $m$. This makes the integral easier to solve.

And we have our integral of the form $\int {\dfrac{{P\left( x \right)}}{{Q\left( x \right)}}} dx$, where we can check for derivative of the function $Q\left( x \right)$

$ \Rightarrow \dfrac{{d\left( {{{\left( {{x^5} + {x^3} + 1} \right)}^3}} \right)}}{{dx}} = 3\left( {5{x^4} + 3{x^2}} \right)$ ; which will not help us in any way to make it easier.

Let’s take out the common from the part $\left( {{x^5} + {x^3} + 1} \right)$, we can write it as:

\[ \Rightarrow \int {\dfrac{{2{x^{12}} + 5{x^9}}}{{{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} dx = \int {\dfrac{{2{x^{12}} + 5{x^9}}}{{{{\left( {{x^5}\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)} \right)}^3}}}} dx = \int {\dfrac{{2{x^{12}} + 5{x^9}}}{{{x^{15}}{{\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}^3}}}} dx\]

Now, we can divide our numerator with ${x^{15}}$, this will give us:

\[ \Rightarrow \int {\dfrac{{2{x^{12}} + 5{x^9}}}{{{x^{15}}{{\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}^3}}}} dx = \int {\dfrac{{2{x^{12 - 15}} + 5{x^{9 - 15}}}}{{{{\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}^3}}}} dx = \int {\dfrac{{2{x^{ - 3}} + 5{x^{ - 6}}}}{{{{\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}^3}}}} dx\]

And we still have our integral of the form$\int {\dfrac{{P\left( x \right)}}{{Q\left( x \right)}}} dx$, where we can check for derivative of the function $Q\left( x \right)$

$ \Rightarrow \dfrac{{d\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}}{{dx}} = 0 + \left( { - 2{x^{ - 2 - 1}}} \right) + \left( { - 5{x^{ - 5 - 1}}} \right) = - 2{x^{ - 3}} - 5{x^{ - 6}} = - P\left( x \right)$

Therefore, we can now apply the substitution method in our integral:

$ \Rightarrow \left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right) = m \Rightarrow d\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right) = dm \Rightarrow dm = - 2{x^{ - 3}} - 5{x^{ - 6}}$

Let’s put this value in our integral:

\[ \Rightarrow \int {\dfrac{{2{x^{ - 3}} + 5{x^{ - 6}}}}{{{{\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}^3}}}} dx = \int {\dfrac{{ - dm}}{{{m^3}}}} \]

As we know that,$\int {{x^a}} dx = \dfrac{{{x^{a + 1}}}}{{a + 1}} + C$, which can be used in our integral

$ \Rightarrow \int {\dfrac{{ - dm}}{{{m^3}}}} = - \dfrac{{{m^{ - 3 + 1}}}}{{\left( { - 3 + 1} \right)}} + C = - \dfrac{{{m^{ - 2}}}}{{ - 2}} + C = \dfrac{1}{{2{m^2}}} + C$

Now, we can again substitute the value of $m$ into the expression as:

$ \Rightarrow \dfrac{1}{{2{m^2}}} + C = \dfrac{1}{{2{{\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}^2}}} + C$

This can be further simplified as:

$ \Rightarrow \dfrac{1}{{2{{\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}^2}}} + C = \dfrac{1}{{2{{\left( {{x^{ - 5}}\left( {{x^5} + {x^3} + 1} \right)} \right)}^2}}} + C = \dfrac{1}{{2{x^{ - 10}}{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C = \dfrac{{{x^{10}}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C$

Thus we got the integral as: $\int {\dfrac{{2{x^{12}} + 5{x^9}}}{{{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} dx = \dfrac{{{x^{10}}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C$

**So, the correct answer is “Option B”.**

**Note:**Try to go step by step with the solution to avoid the complications. Be careful with the substitution in the integral. Always take care of the signs after substitutions. Here we used $\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right) = m$. Notice that the integral on the left is expressed in terms of the variable$x$. The integral on the right is in terms of$m$.Do not forget to express the final answer in terms of the original variable$x$.

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