If $\displaystyle\int_0^1(x^{21}+x^{14}+x^7)(2x^{14}+3x^7+6)^{1/7}\,dx=\dfrac{1}{l}(11)^{m/n}$, where $l,m,n\in\mathbb{N}$ and $m,n$ are coprime, then $l+m+n$ is equal to ___.
Step-by-Step Solution
Key Concept: Let $t=x^7$: integral $=\frac{1}{7}\int_0^1 t^{1/7}(t^2+t+1)(2t^2+3t+6)^{1/7}dt$. Let $u=(2t^3+3t^2+6t)^{1/7}$: $7u^6du=6(t^2+t+1)dt$.
Step 1: Identify a suitable substitution to simplify the integrand.
The powers of $x$ in the expression $(x^{21}+x^{14}+x^7)$ are multiples of 7. The powers of $x$ inside $(2x^{14}+3x^7+6)$ are also multiples of 7 (or constants). This suggests a substitution involving $x^7$. Let $y = x^7$.
Step 2: Transform the integral using the substitution.
If $y = x^7$, then $dy = 7x^6 dx$. We need to express $dx$ in terms of $dy$ and $y$.
Since $x = y^{1/7}$, we have $x^6 = (y^{1/7})^6 = y^{6/7}$.
So, $dx = \frac{dy}{7x^6} = \frac{dy}{7y^{6/7}}$.
Now, let's change the limits of integration:
When $x=0$, $y=0^7=0$.
When $x=1$, $y=1^7=1$.
Substitute $y=x^7$ into the original integral:
$$I = \int_0^1 (y^3+y^2+y)(2y^2+3y+6)^{1/7} \left(\frac{dy}{7y^{6/7}}\right)$$
Factor out $y$ from the first term $(y^3+y^2+y)$ and bring $y^{6/7}$ into the first parenthesis:
$$I = \frac{1}{7} \int_0^1 (y^3+y^2+y)y^{-6/7}(2y^2+3y+6)^{1/7}\,dy$$
$$I = \frac{1}{7} \int_0^1 (y^{3-6/7}+y^{2-6/7}+y^{1-6/7})(2y^2+3y+6)^{1/7}\,dy$$
$$I = \frac{1}{7} \int_0^1 (y^{15/7}+y^{8/7}+y^{1/7})(2y^2+3y+6)^{1/7}\,dy$$
Now, factor out $y^{1/7}$ from the first bracket:
$$I = \frac{1}{7} \int_0^1 y^{1/7}(y^{14/7}+y^{7/7}+1)(2y^2+3y+6)^{1/7}\,dy$$
$$I = \frac{1}{7} \int_0^1 y^{1/7}(y^2+y+1)(2y^2+3y+6)^{1/7}\,dy$$
Bring $y^{1/7}$ inside the term raised to the power $1/7$: $y^{1/7} = (y)^{1/7}$.
$$I = \frac{1}{7} \int_0^1 (y^2+y+1) (y(2y^2+3y+6))^{1/7}\,dy$$
$$I = \frac{1}{7} \int_0^1 (y^2+y+1) (2y^3+3y^2+6y)^{1/7}\,dy$$
Step 3: Perform a second substitution to evaluate the integral.
Let $u = 2y^3+3y^2+6y$.
Then, $du = (6y^2+6y+6)dy = 6(y^2+y+1)dy$.
So, $(y^2+y+1)dy = \frac{1}{6}du$.
Change the limits of integration for $u$:
When $y=0$, $u = 2(0)^3+3(0)^2+6(0) = 0$.
When $y=1$, $u = 2(1)^3+3(1)^2+6(1) = 2+3+6 = 11$.
Substitute these into the integral:
$$I = \frac{1}{7} \int_0^{11} (u)^{1/7} \left(\frac{1}{6}\right)\,du$$
$$I = \frac{1}{42} \int_0^{11} u^{1/7}\,du$$
Now, integrate with respect to $u$:
$$I = \frac{1}{42} \left[ \frac{u^{1/7+1}}{1/7+1} \right]_0^{11}$$
$$I = \frac{1}{42} \left[ \frac{u^{8/7}}{8/7} \right]_0^{11}$$
$$I = \frac{1}{42} \left[ \frac{7}{8} u^{8/7} \right]_0^{11}$$
$$I = \frac{1}{42} \cdot \frac{7}{8} (11^{8/7} - 0^{8/7})$$
$$I = \frac{1}{6 \cdot 8} (11^{8/7})$$
$$I = \frac{1}{48} (11^{8/7})$$
Step 4: Compare the result with the given form and identify $l, m, n$.
The calculated value of the integral is $\dfrac{1}{48}(11)^{8/7}$.
The problem states that the integral is equal to $\dfrac{1}{l}(11)^{m/n}$.
Comparing the two forms, we get:
$l = 48$
$m = 8$
$n = 7$
We check that $m=8$ and $n=7$ are coprime natural numbers, which they are.
Step 5: Calculate $l+m+n$.
$l+m+n = 48+8+7 = 63$.
The final answer is $\boxed{63}$.
Correct Answer: 63