Key Concepts and Formulas

• Substitution Method for Integration: If $\int f(g(x))g'(x) dx$ can be written, substitute $u = g(x)$ and $du = g'(x)dx$, so the integral becomes $\int f(u) du$.
• Power Rule for Integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$.
• Definite Integral: $\int_a^b f(x) dx = F(b) - F(a)$, where $F(x)$ is the antiderivative of $f(x)$.

Step-by-Step Solution

Step 1: Rewrite the integral

We are given the integral $I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}}$ We want to simplify this integral using a substitution.

Step 2: Perform the substitution

Let's try the substitution $t = \frac{x-11}{x+15}$. Then, we need to find $dt$ in terms of $dx$. $t = \frac{x-11}{x+15}$ Differentiating both sides with respect to $x$, we get: $\frac{dt}{dx} = \frac{(x+15)(1) - (x-11)(1)}{(x+15)^2} = \frac{x+15 - x + 11}{(x+15)^2} = \frac{26}{(x+15)^2}$ So, $dt = \frac{26}{(x+15)^2} dx$, which implies $dx = \frac{(x+15)^2}{26} dt$.

Now, we rewrite the integral in terms of $t$. We have $x-11 = t(x+15)$, so $I(x) = \int \frac{1}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}} dx = \int \frac{1}{(t(x+15))^{\frac{11}{13}}(x+15)^{\frac{15}{13}}} dx = \int \frac{1}{t^{\frac{11}{13}}(x+15)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}} dx = \int \frac{1}{t^{\frac{11}{13}}(x+15)^{\frac{26}{13}}} dx = \int \frac{1}{t^{\frac{11}{13}}(x+15)^2} dx$ Substituting $dx = \frac{(x+15)^2}{26} dt$, we get $I(x) = \int \frac{1}{t^{\frac{11}{13}}(x+15)^2} \frac{(x+15)^2}{26} dt = \int \frac{1}{t^{\frac{11}{13}}} \frac{1}{26} dt = \frac{1}{26} \int t^{-\frac{11}{13}} dt$

Step 3: Evaluate the integral

Now, we can integrate with respect to $t$ using the power rule: $I(x) = \frac{1}{26} \int t^{-\frac{11}{13}} dt = \frac{1}{26} \frac{t^{-\frac{11}{13}+1}}{-\frac{11}{13}+1} + C = \frac{1}{26} \frac{t^{\frac{2}{13}}}{\frac{2}{13}} + C = \frac{1}{26} \cdot \frac{13}{2} t^{\frac{2}{13}} + C = \frac{1}{4} t^{\frac{2}{13}} + C$ Substituting back $t = \frac{x-11}{x+15}$, we get $I(x) = \frac{1}{4} \left(\frac{x-11}{x+15}\right)^{\frac{2}{13}} + C$

Step 4: Calculate I(37) - I(24)

We are given that $I(37) - I(24) = \frac{1}{4}\left(\frac{1}{b^{\frac{1}{13}}} - \frac{1}{c^{\frac{1}{13}}}\right)$. $I(37) = \frac{1}{4} \left(\frac{37-11}{37+15}\right)^{\frac{2}{13}} + C = \frac{1}{4} \left(\frac{26}{52}\right)^{\frac{2}{13}} + C = \frac{1}{4} \left(\frac{1}{2}\right)^{\frac{2}{13}} + C$ $I(24) = \frac{1}{4} \left(\frac{24-11}{24+15}\right)^{\frac{2}{13}} + C = \frac{1}{4} \left(\frac{13}{39}\right)^{\frac{2}{13}} + C = \frac{1}{4} \left(\frac{1}{3}\right)^{\frac{2}{13}} + C$ Therefore, $I(37) - I(24) = \frac{1}{4} \left(\frac{1}{2}\right)^{\frac{2}{13}} - \frac{1}{4} \left(\frac{1}{3}\right)^{\frac{2}{13}} = \frac{1}{4} \left(\frac{1}{2^{\frac{2}{13}}} - \frac{1}{3^{\frac{2}{13}}}\right) = \frac{1}{4} \left(\frac{1}{(2^2)^{\frac{1}{13}}} - \frac{1}{(3^2)^{\frac{1}{13}}}\right) = \frac{1}{4} \left(\frac{1}{4^{\frac{1}{13}}} - \frac{1}{9^{\frac{1}{13}}}\right)$

Step 5: Find b and c

Comparing $I(37) - I(24) = \frac{1}{4}\left(\frac{1}{b^{\frac{1}{13}}} - \frac{1}{c^{\frac{1}{13}}}\right)$ with $I(37) - I(24) = \frac{1}{4} \left(\frac{1}{4^{\frac{1}{13}}} - \frac{1}{9^{\frac{1}{13}}}\right)$, we get $b = 4$ and $c = 9$.

Step 6: Calculate 3(b+c)

$3(b+c) = 3(4+9) = 3(13) = 39$

Common Mistakes & Tips

• Algebra Errors: Double-check algebraic manipulations, especially when dealing with fractions and exponents.
• Substitution Complications: Choose substitutions wisely. The goal is to simplify the integral, not make it more complicated.
• Forgetting the Constant of Integration: Remember to add "+ C" after evaluating indefinite integrals. However, since this is a definite integral difference, the C cancels out.

Summary

We solved the indefinite integral using substitution and then evaluated the definite integral difference $I(37) - I(24)$. By comparing the result with the given expression, we found the values of $b$ and $c$, and finally calculated $3(b+c)$, which is 39.

Final Answer

The final answer is \boxed{39}, which corresponds to option (A).