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

Find the integral 2x12+5x9(x5+x3+1)3dx\int {\dfrac{{2{x^{12}} + 5{x^9}}}{{{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} dx
(a) x5(x5+x3+1)2+C\dfrac{{ - {x^5}}}{{{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C
(b) x102(x5+x3+1)2+C\dfrac{{{x^{10}}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C
(c) x52(x5+x3+1)2+C\dfrac{{{x^5}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C
(d) x102(x5+x3+1)2+C\dfrac{{ - {x^{ - 10}}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C

Explanation

Solution

Start with trying to take common x5{x^5}from the denominator term(x5+x3+1)\left( {{x^5} + {x^3} + 1} \right). This will give you x15{x^{15}}in the denominator and then you can divide the numerator withx15{x^{15}}. This will give you an expression(1+x2+x5)\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 integrationf(g(x))g(x)dx=f(u)du,where,u=g(x)\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 mm equal to the function and rewrite the integral in terms of the new variable mm. This makes the integral easier to solve.

And we have our integral of the form P(x)Q(x)dx\int {\dfrac{{P\left( x \right)}}{{Q\left( x \right)}}} dx, where we can check for derivative of the function Q(x)Q\left( x \right)
d((x5+x3+1)3)dx=3(5x4+3x2)\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 (x5+x3+1)\left( {{x^5} + {x^3} + 1} \right), we can write it as:
2x12+5x9(x5+x3+1)3dx=2x12+5x9(x5(1+x2+x5))3dx=2x12+5x9x15(1+x2+x5)3dx\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 x15{x^{15}}, this will give us:
2x12+5x9x15(1+x2+x5)3dx=2x1215+5x915(1+x2+x5)3dx=2x3+5x6(1+x2+x5)3dx\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 formP(x)Q(x)dx\int {\dfrac{{P\left( x \right)}}{{Q\left( x \right)}}} dx, where we can check for derivative of the function Q(x)Q\left( x \right)
d(1+x2+x5)dx=0+(2x21)+(5x51)=2x35x6=P(x)\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:
(1+x2+x5)=md(1+x2+x5)=dmdm=2x35x6\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:
2x3+5x6(1+x2+x5)3dx=dmm3\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,xadx=xa+1a+1+C\int {{x^a}} dx = \dfrac{{{x^{a + 1}}}}{{a + 1}} + C, which can be used in our integral
dmm3=m3+1(3+1)+C=m22+C=12m2+C\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 mm into the expression as:
12m2+C=12(1+x2+x5)2+C\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:
12(1+x2+x5)2+C=12(x5(x5+x3+1))2+C=12x10(x5+x3+1)2+C=x102(x5+x3+1)2+C\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: 2x12+5x9(x5+x3+1)3dx=x102(x5+x3+1)2+C\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 (1+x2+x5)=m\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right) = m. Notice that the integral on the left is expressed in terms of the variablexx. The integral on the right is in terms ofmm.Do not forget to express the final answer in terms of the original variablexx.