Question

Let $I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}}$. If

$I(37) - I(24) = \frac{1}{4} \left( \frac{1}{b^{\frac{1}{13}}} - \frac{1}{c^{\frac{1}{13}}} \right)$, $b, c \in \mathbb{N}$, then $3(b+c)$ is equal to

Answer 1

39

Answer 2

To evaluate the integral $I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}}$, we first observe the powers in the denominator. The sum of the powers is $\frac{11}{13} + \frac{15}{13} = \frac{26}{13} = 2$. This suggests a substitution of the form $u = \frac{x-a}{x-b}$.

Let's rewrite the integrand by factoring out $(x+15)^2$ from the denominator: $I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{11}{13}}(x+15)^{\frac{4}{13}}} = \int \frac{dx}{\left((x-11)(x+15)\right)^{\frac{11}{13}}(x+15)^{\frac{4}{13}}}$ This is not the most straightforward way. Instead, let's divide the numerator and denominator by $(x+15)^2$: $I(x) = \int \frac{\frac{1}{(x+15)^2} dx}{\frac{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}}{(x+15)^2}} = \int \frac{\frac{1}{(x+15)^2} dx}{\left(\frac{x-11}{x+15}\right)^{\frac{11}{13}}}$

Now, let $u = \frac{x-11}{x+15}$. Differentiating $u$ with respect to $x$: $du = \frac{d}{dx}\left(\frac{x-11}{x+15}\right) dx = \frac{(x+15)(1) - (x-11)(1)}{(x+15)^2} dx$ $du = \frac{x+15-x+11}{(x+15)^2} dx = \frac{26}{(x+15)^2} dx$ From this, we have $\frac{dx}{(x+15)^2} = \frac{1}{26} du$.

Substitute $u$ and $du$ into the integral: $I(x) = \int \frac{1}{u^{\frac{11}{13}}} \cdot \frac{1}{26} du = \frac{1}{26} \int u^{-\frac{11}{13}} du$ Now, integrate $u^{-\frac{11}{13}}$: $\int u^{-\frac{11}{13}} du = \frac{u^{-\frac{11}{13}+1}}{-\frac{11}{13}+1} + C = \frac{u^{\frac{2}{13}}}{\frac{2}{13}} + C = \frac{13}{2} u^{\frac{2}{13}} + C$ Substitute this back into the expression for $I(x)$: $I(x) = \frac{1}{26} \cdot \frac{13}{2} u^{\frac{2}{13}} + C = \frac{1}{4} u^{\frac{2}{13}} + C$ Finally, substitute back $u = \frac{x-11}{x+15}$: $I(x) = \frac{1}{4} \left(\frac{x-11}{x+15}\right)^{\frac{2}{13}} + C$

Now we need to evaluate $I(37) - I(24)$. For definite integrals, the constant $C$ cancels out. $I(37) = \frac{1}{4} \left(\frac{37-11}{37+15}\right)^{\frac{2}{13}} = \frac{1}{4} \left(\frac{26}{52}\right)^{\frac{2}{13}} = \frac{1}{4} \left(\frac{1}{2}\right)^{\frac{2}{13}}$ $I(24) = \frac{1}{4} \left(\frac{24-11}{24+15}\right)^{\frac{2}{13}} = \frac{1}{4} \left(\frac{13}{39}\right)^{\frac{2}{13}} = \frac{1}{4} \left(\frac{1}{3}\right)^{\frac{2}{13}}$ Subtracting $I(24)$ from $I(37)$: $I(37) - I(24) = \frac{1}{4} \left[ \left(\frac{1}{2}\right)^{\frac{2}{13}} - \left(\frac{1}{3}\right)^{\frac{2}{13}} \right]$ $I(37) - I(24) = \frac{1}{4} \left[ \frac{1}{2^{\frac{2}{13}}} - \frac{1}{3^{\frac{2}{13}}} \right]$ We can rewrite the terms in the form $k^{\frac{1}{13}}$: $2^{\frac{2}{13}} = (2^2)^{\frac{1}{13}} = 4^{\frac{1}{13}}$ $3^{\frac{2}{13}} = (3^2)^{\frac{1}{13}} = 9^{\frac{1}{13}}$ So, $I(37) - I(24) = \frac{1}{4} \left[ \frac{1}{4^{\frac{1}{13}}} - \frac{1}{9^{\frac{1}{13}}} \right]$ The problem states that $I(37) - I(24) = \frac{1}{4} \left( \frac{1}{b^{\frac{1}{13}}} - \frac{1}{c^{\frac{1}{13}}} \right)$. Comparing the two expressions, we get $b=4$ and $c=9$. Both $b$ and $c$ are natural numbers, as required.

Finally, we need to find the value of $3(b+c)$: $3(b+c) = 3(4+9) = 3(13) = 39$

The final answer is $\boxed{39}$.

Explanation of the solution:

1. Identify the integral type: The integral is of the form $\int \frac{dx}{(x-a)^m (x-b)^n}$ where $m+n$ is an integer ($11/13 + 15/13 = 2$). This suggests a substitution $u = \frac{x-a}{x-b}$.
2. Choose substitution: Let $u = \frac{x-11}{x+15}$.
3. Calculate $du$: Find the derivative of $u$ with respect to $x$, which is $du = \frac{26}{(x+15)^2} dx$.
4. Rewrite integrand: Express the original integrand in terms of $u$ and $du$. This involves dividing the numerator and denominator by $(x+15)^2$.
5. Integrate: The integral transforms into $\frac{1}{26} \int u^{-\frac{11}{13}} du$, which evaluates to $\frac{1}{4} u^{\frac{2}{13}}$.
6. Substitute back: Replace $u$ with $\frac{x-11}{x+15}$ to get $I(x) = \frac{1}{4} \left(\frac{x-11}{x+15}\right)^{\frac{2}{13}}$.
7. Evaluate definite integral: Calculate $I(37)$ and $I(24)$ and find their difference.
8. Compare and solve: Match the result with the given form $\frac{1}{4} \left( \frac{1}{b^{\frac{1}{13}}} - \frac{1}{c^{\frac{1}{13}}} \right)$ to find $b=4$ and $c=9$.
9. Calculate final expression: Compute $3(b+c)$.

The final answer is $\boxed{39}$.