Question

The integral $\int {({x^{7m}} + {x^{2m}} + {x^m}){{(2{x^{6m}} + 7{x^m} + 14)}^{\dfrac{1}{m}}}} .dx$ equal to ____

A.$\dfrac{{{{({x^{7m}} + {x^{2m}} + {x^m})}^{m + 1}}}}{{m + 1}} + c$
B.$\dfrac{1}{{14(m + 1)}}.{\left( {2{x^{7m}} + 7{x^{2m}} + 14{x^m}} \right)^{\dfrac{{1 + m}}{m}}} + c$
C.${\left( {2{x^{7m}} + 7{x^{2m}} + 14{x^m}} \right)^{\dfrac{{1 + m}}{m}}} + c$
D.$\dfrac{1}{{14(m + 1)}}.\left( {2{x^{7m}} + 7{x^{2m}} + 14{x^m}} \right) + c$

Answer 1

Hint : We know the integration formula, $\int {{x^n}.dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$. We simplify the given integral then by using the substitution method we can solve this. The substitution method is used when an integral contains some function and its derivative. After solving by substitution method we keep the answer in terms of ‘x’ only.

Complete step-by-step answer :
Given, $\int {({x^{7m}} + {x^{2m}} + {x^m}){{(2{x^{6m}} + 7{x^m} + 14)}^{\dfrac{1}{m}}}} .dx$
Now multiplying and divide by ${x^m}$inside the second term, we have:
$\Rightarrow \int {({x^{7m}} + {x^{2m}} + {x^m}){{\left( {\dfrac{{{x^m}}}{{{x^m}}} \times (2{x^{6m}} + 7{x^m} + 14)} \right)}^{\dfrac{1}{m}}}} .dx$
Multiplying we have,
$\Rightarrow \int {({x^{7m}} + {x^{2m}} + {x^m}){{\left( {\dfrac{{(2{x^{6m}} \times {x^m} + 7{x^m} \times {x^m} + 14 \times {x^m})}}{{{x^m}}}} \right)}^{\dfrac{1}{m}}}} .dx$
Since the bases are same then we add it’s powers, ${a^m}.{a^n} = {a^{m + n}}$ and we know ${\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}$, then we get,
$\Rightarrow \int {({x^{7m}} + {x^{2m}} + {x^m})\dfrac{{{{(2{x^{6m + m}} + 7{x^{m + m}} + 14{x^m})}^{\dfrac{1}{m}}}}}{{{{({x^m})}^{\dfrac{1}{m}}}}}} .dx$
$\Rightarrow \int {({x^{7m}} + {x^{2m}} + {x^m})\dfrac{{{{(2{x^{7m}} + 7{x^{2m}} + 14{x^m})}^{\dfrac{1}{m}}}}}{{(x)}}} .dx$
Rearranging we get,
$\Rightarrow \int {\dfrac{{({x^{7m}} + {x^{2m}} + {x^m})}}{x}{{(2{x^{7m}} + 7{x^{2m}} + 14{x^m})}^{\dfrac{1}{m}}}} .dx$
$\Rightarrow \int {{x^{ - 1}}({x^{7m}} + {x^{2m}} + {x^m}){{(2{x^{7m}} + 7{x^{2m}} + 14{x^m})}^{\dfrac{1}{m}}}} .dx$
Again we use ${a^m}.{a^n} = {a^{m + n}}$, we get:
$\Rightarrow \int {({x^{7m - 1}} + {x^{2m - 1}} + {x^{m - 1}}){{(2{x^{7m}} + 7{x^{2m}} + 14{x^m})}^{\dfrac{1}{m}}}} .dx{\text{ - - - - - - - (1)}}$
We apply the substitution method.
That is, let $2{x^{7m}} + 7{x^{2m}} + 14{x^m} = t{\text{ - - - - - - - (2)}}$
Differentiating with respect to ‘x’ we have,
$\Rightarrow (14m{x^{7m - 1}} + 14m{x^{2m - 1}} + 14m{x^{m - 1}})dx = dt$
Taking 14m as common term,
$\Rightarrow 14m({x^{7m - 1}} + {x^{2m - 1}} + {x^{m - 1}})dx = dt$
Divided by 14m on the both side we have,
$\Rightarrow ({x^{7m - 1}} + {x^{2m - 1}} + {x^{m - 1}})dx = \dfrac{{dt}}{{14m}}{\text{ - - - - - - (3)}}$
From (1) we have
$\Rightarrow \int {({x^{7m - 1}} + {x^{2m - 1}} + {x^{m - 1}}){{(2{x^{7m}} + 7{x^{2m}} + 14{x^m})}^{\dfrac{1}{m}}}} .dx$
Using equation (3) we have,
$\Rightarrow \int {\dfrac{{dt}}{{14m}}{{(t)}^{\dfrac{1}{m}}}}$
$\Rightarrow \int {\dfrac{1}{{14m}}{{(t)}^{\dfrac{1}{m}}}} .dt$
Taking constant outside we have,
$\Rightarrow \dfrac{1}{{14m}}\int {{{(t)}^{\dfrac{1}{m}}}} .dt$
Integrating with respect to ‘t’, we get:
$\Rightarrow \dfrac{1}{{14m}} \times \left( {\dfrac{{{t^{\dfrac{1}{m} + 1}}}}{{\dfrac{1}{m} + 1}}} \right) + c$
Where, c is the integration constant and taking L.C.M. we have:
$\Rightarrow \dfrac{1}{{14m}} \times \left( {\dfrac{{{t^{\dfrac{{1 + m}}{m}}}}}{{\dfrac{{1 + m}}{m}}}} \right) + c$
Rearranging we have,
$\Rightarrow \dfrac{m}{{14m(m + 1)}}.{t^{\dfrac{{1 + m}}{m}}} + c$
Cancelling m we have,
$\Rightarrow \dfrac{1}{{14(m + 1)}}.{t^{\dfrac{{1 + m}}{m}}} + c$
Now we need to keep the answer in terms of ‘x’ only,
From equation (2) we have $2{x^{7m}} + 7{x^{2m}} + 14{x^m} = t$. Now substituting value of ‘t’ we get:
$\Rightarrow \dfrac{1}{{14(m + 1)}}.{\left( {2{x^{7m}} + 7{x^{2m}} + 14{x^m}} \right)^{\dfrac{{1 + m}}{m}}} + c$
So, the correct answer is “Option B”.

Note : We know that in indefinite integral we have integral constant but in definite integral we do not have an integral constant. By using the substitution method we can easily solve the integral as done above. In case of definite integral while using substitution methods limits also change, careful about that part.