Question

For any natural number m, \[{{\int{\left( {{x}^{7m}}+{{x}^{2m}}+{{x}^{m}} \right)\left( 2{{x}^{6m}}+7{{x}^{m}}+14 \right)}}^{\dfrac{1}{m}}}dx\] where \[x>0\] equals;
A. \[\dfrac{{{\left( 7{{x}^{7m}}+2{{x}^{2m}}+14{{x}^{m}} \right)}^{\dfrac{m+1}{m}}}}{14\left( m+1 \right)}+C\]
B. \[\dfrac{{{\left( 2{{x}^{7m}}+14{{x}^{2m}}+7{{x}^{m}} \right)}^{\dfrac{m+1}{m}}}}{14\left( m+1 \right)}+C\]
C. \[\dfrac{{{\left( 2{{x}^{7m}}+7{{x}^{2m}}+14{{x}^{m}} \right)}^{\dfrac{m+1}{m}}}}{14\left( m+1 \right)}+C\]
D. \[\dfrac{{{\left( 7{{x}^{7m}}+2{{x}^{2m}}+{{x}^{m}} \right)}^{\dfrac{m+1}{m}}}}{14\left( m+1 \right)}+C\]

Answer 1

Hint: In order to integrate the given expression, first we need to rewrite the given expression as \[{{\int{\left( {{x}^{7m}}+{{x}^{2m}}+{{x}^{m}} \right)\left( \left( 2{{x}^{6m}}+7{{x}^{m}}+14 \right)\cdot \dfrac{{{x}^{m}}}{{{x}^{m}}} \right)}}^{\dfrac{1}{m}}}dx\] . Then applying the laws of exponents and power i.e. \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\] and \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\] , we will simplifying the expression. Later to integrate the given expression, we need to substitute \[\left( 2{{x}^{7m}}+7{{x}^{2m}}+14{{x}^{m}} \right)=t\] and then differentiating it with respect to ‘x’ and substituting the values and integrating by applying the rules of integration. In this way we will get the required answer.
Formula used:
If ‘a’ is the positive rational number and ‘m’ and ‘n’ are the given rational exponent either positive exponent or negative exponent, then
 \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\]
 \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\]

Complete step-by-step answer:
We have given that,
 \[{{\int{\left( {{x}^{7m}}+{{x}^{2m}}+{{x}^{m}} \right)\left( 2{{x}^{6m}}+7{{x}^{m}}+14 \right)}}^{\dfrac{1}{m}}}dx\] , where \[x>0\] and ‘m’ is any natural number.
Let I be the given integral.
Therefore,
 \[\Rightarrow I={{\int{\left( {{x}^{7m}}+{{x}^{2m}}+{{x}^{m}} \right)\left( 2{{x}^{6m}}+7{{x}^{m}}+14 \right)}}^{\dfrac{1}{m}}}dx\]
It can be rewritten as,
 \[\Rightarrow I={{\int{\left( {{x}^{7m}}+{{x}^{2m}}+{{x}^{m}} \right)\left( \left( 2{{x}^{6m}}+7{{x}^{m}}+14 \right)\cdot \dfrac{{{x}^{m}}}{{{x}^{m}}} \right)}}^{\dfrac{1}{m}}}dx\]
Simplifying the above, we will get
Using the laws of exponents i.e. \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\] and \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\]
 \[\Rightarrow I={{\int{\left( {{x}^{7m-1}}+{{x}^{2m-1}}+{{x}^{m-1}} \right)\left( \left( 2{{x}^{7m}}+7{{x}^{2m}}+14{{x}^{m}} \right) \right)}}^{\dfrac{1}{m}}}dx\]
Now,
Substituting the \[\left( 2{{x}^{7m}}+7{{x}^{2m}}+14{{x}^{m}} \right)=t\]
We have,
 \[\left( 2{{x}^{7m}}+7{{x}^{2m}}+14{{x}^{m}} \right)=t\]
Differentiating it with respect to ‘x’,
 \[\left( 14m{{x}^{7m-1}}+14m{{x}^{2m-1}}+14m{{x}^{m-1}} \right)dx=dt\]
Or,
 \[\left( {{x}^{7m-1}}+{{x}^{2m-1}}+{{x}^{m-1}} \right)dx=\dfrac{dt}{14m}\]
Substituting this value in the given above integral,
We will get,
 \[\Rightarrow I=\dfrac{1}{14m}\int{{{t}^{\dfrac{1}{m}}}dt}\]
Applying the rules of integration;
 \[\Rightarrow I=\dfrac{1}{14m}\int{{{t}^{\dfrac{1}{m}}}dt}=\dfrac{{{t}^{\dfrac{1}{m}+1}}}{14\left( m+1 \right)}+C\]
Undo the substitution i.e. \[\left( 2{{x}^{7m}}+7{{x}^{2m}}+14{{x}^{m}} \right)=t\]
 \[\Rightarrow I=\dfrac{{{\left( 2{{x}^{7m}}+7{{x}^{2m}}+14{{x}^{m}} \right)}^{\dfrac{1}{m}+1}}}{14\left( m+1 \right)}+C\]
Therefore,
 \[\Rightarrow {{\int{\left( {{x}^{7m}}+{{x}^{2m}}+{{x}^{m}} \right)\left( 2{{x}^{6m}}+7{{x}^{m}}+14 \right)}}^{\dfrac{1}{m}}}dx=\dfrac{{{\left( 2{{x}^{7m}}+7{{x}^{2m}}+14{{x}^{m}} \right)}^{\dfrac{1}{m}+1}}}{14\left( m+1 \right)}+C\]
Hence, the option (c ) is the correct answer.
So, the correct answer is “Option C”.

Note: While solving these types of questions, students should remember all the laws of exponents and powers as then we will be able to solve the question easily. Students need to know that the integration by substitution also known as u-substitution or change of variables, is a method which is used for evaluating integrals or anti-derivatives. Students need to do the calculation part very carefully to avoid making any type of error.