How do you integrate $\int{36{{x}^{3}}{{\left( 3{{x}^{4}}+3 \right)}^{5}}}$ using substitution?
(a) Using integrating properties
(b) Using differentiation properties
(c) a and b both
(d) None of these
Answer
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Hint: We are to find the integration of $\int{36{{x}^{3}}{{\left( 3{{x}^{4}}+3 \right)}^{5}}}dx$. We will start with substituting $u=3{{x}^{4}}+3$to get the solution. Now, using the integration properties of $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$, we can easily simplify the problem and get the solution. Putting the value of u back will give us the final result.
Complete step by step solution:
According to the question, we are to find the integration of $\int{36{{x}^{3}}{{\left( 3{{x}^{4}}+3 \right)}^{5}}}dx$.
We will use to integrating property, $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ for n = -1,
We can do it by substitution. We start with, $u=3{{x}^{4}}+3$
Thus, we have, \[du=3.4{{x}^{3}}dx=12{{x}^{3}}dx\]
Dividing both sides by 12, we get, ${{x}^{3}}dx=\dfrac{du}{12}$
So, $\int{36{{x}^{3}}{{\left( 3{{x}^{4}}+3 \right)}^{5}}}dx$ can be written as,
$\Rightarrow \dfrac{36}{12}\int{{{u}^{5}}du}=3\int{{{u}^{5}}du}$
Again using, $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$,we get, $\int{{{u}^{5}}dx}=\dfrac{{{u}^{6}}}{6}$
Thus, it can be written as,
$\Rightarrow 3.\dfrac{{{u}^{6}}}{6}+C$ , where C is a constant.
Now, simplifying,
$\Rightarrow \dfrac{{{u}^{6}}}{2}+C$
Putting the value of u back in the given equation, we are getting now,
$\Rightarrow \dfrac{{{\left( 3{{x}^{4}}+3 \right)}^{6}}}{2}+C$
We can also write it in a more simplified form. We will take 3 common from $3{{x}^{4}}+3$ to write the more simplified form.
$\Rightarrow \dfrac{{{3}^{6}}{{\left( {{x}^{4}}+1 \right)}^{6}}}{2}+C$
Again, ${{3}^{6}}$ can be written as ${{3}^{6}}=729$. Thus, we have the end solution as,
$\Rightarrow \dfrac{729{{\left( {{x}^{4}}+1 \right)}^{6}}}{2}+C$
So, the correct answer is “Option C”.
Note: In this problem, we have integrated the given term using one single property $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$. But there are many properties which can be used to find the solutions. The other properties can be used to get another simplified solution of the same problem. The properties conclude the absolute values and different terms to deal with.
Complete step by step solution:
According to the question, we are to find the integration of $\int{36{{x}^{3}}{{\left( 3{{x}^{4}}+3 \right)}^{5}}}dx$.
We will use to integrating property, $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ for n = -1,
We can do it by substitution. We start with, $u=3{{x}^{4}}+3$
Thus, we have, \[du=3.4{{x}^{3}}dx=12{{x}^{3}}dx\]
Dividing both sides by 12, we get, ${{x}^{3}}dx=\dfrac{du}{12}$
So, $\int{36{{x}^{3}}{{\left( 3{{x}^{4}}+3 \right)}^{5}}}dx$ can be written as,
$\Rightarrow \dfrac{36}{12}\int{{{u}^{5}}du}=3\int{{{u}^{5}}du}$
Again using, $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$,we get, $\int{{{u}^{5}}dx}=\dfrac{{{u}^{6}}}{6}$
Thus, it can be written as,
$\Rightarrow 3.\dfrac{{{u}^{6}}}{6}+C$ , where C is a constant.
Now, simplifying,
$\Rightarrow \dfrac{{{u}^{6}}}{2}+C$
Putting the value of u back in the given equation, we are getting now,
$\Rightarrow \dfrac{{{\left( 3{{x}^{4}}+3 \right)}^{6}}}{2}+C$
We can also write it in a more simplified form. We will take 3 common from $3{{x}^{4}}+3$ to write the more simplified form.
$\Rightarrow \dfrac{{{3}^{6}}{{\left( {{x}^{4}}+1 \right)}^{6}}}{2}+C$
Again, ${{3}^{6}}$ can be written as ${{3}^{6}}=729$. Thus, we have the end solution as,
$\Rightarrow \dfrac{729{{\left( {{x}^{4}}+1 \right)}^{6}}}{2}+C$
So, the correct answer is “Option C”.
Note: In this problem, we have integrated the given term using one single property $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$. But there are many properties which can be used to find the solutions. The other properties can be used to get another simplified solution of the same problem. The properties conclude the absolute values and different terms to deal with.
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