Find the integral $\int {\dfrac{{2{x^{12}} + 5{x^9}}}{{{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} dx$
(a) $\dfrac{{ - {x^5}}}{{{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C$
(b) $\dfrac{{{x^{10}}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C$
(c) $\dfrac{{{x^5}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C$
(d) $\dfrac{{ - {x^{ - 10}}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C$
Answer
660.3k+ views
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$x$.
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$x$.
Recently Updated Pages
Which of the following graphs shows the variation of class 12 physics CBSE

Draw a labelled diagram of the human male reproductive class 12 biology CBSE

Describe the experiment to compare the emf of two cells class 12 physics CBSE

What is standard hydrogen electrode

What is conventional current and electric current class 12 physics CBSE

2Bromopentane is treated with an alcoholic KOH solution class 12 chemistry CBSE

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

