Key Concepts and Formulas

• Substitution method for integration: $\int f(g(x))g'(x) dx = \int f(u) du$, where $u = g(x)$.
• Properties of exponents: $a^m a^n = a^{m+n}$, $\sqrt[n]{a} = a^{1/n}$, and $(a^m)^n = a^{mn}$.
• Algebraic manipulation to simplify expressions.

Step-by-Step Solution

Step 1: Rewrite the integral. The given integral is $I = \int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} dx$ We want to simplify the expression inside the integral.

Step 2: Perform the substitution $t = \frac{x+3}{x-1}$. Let $t = \frac{x+3}{x-1}$. Then, $t(x-1) = x+3$, which implies $tx - t = x+3$. Solving for $x$, we get $tx - x = t+3$, so $x(t-1) = t+3$, and thus $x = \frac{t+3}{t-1}$.

Step 3: Find $dx$ in terms of $dt$. Differentiating $x = \frac{t+3}{t-1}$ with respect to $t$, we get $\frac{dx}{dt} = \frac{(t-1)(1) - (t+3)(1)}{(t-1)^2} = \frac{t-1-t-3}{(t-1)^2} = \frac{-4}{(t-1)^2}$ Therefore, $dx = \frac{-4}{(t-1)^2} dt$.

Step 4: Rewrite $(x-1)^4(x+3)^6$ in terms of $t$. First, consider $(x-1)^4 (x+3)^6 = (x-1)^4 (x+3)^4 (x+3)^2 = [(x-1)(x+3)]^4 (x+3)^2$. We have $x+3 = t(x-1)$, so $(x-1)^4 (x+3)^6 = (x-1)^4 [t(x-1)]^6 = (x-1)^4 t^6 (x-1)^6 = t^6 (x-1)^{10}$. Also note $(x-1) = \frac{t+3}{t-1} - 1 = \frac{t+3 - (t-1)}{t-1} = \frac{4}{t-1}$. Therefore, $(x-1)^4(x+3)^6 = t^6\left(\frac{4}{t-1}\right)^{10} = t^6 \frac{4^{10}}{(t-1)^{10}}$. Then, $\sqrt[5]{(x-1)^4(x+3)^6} = \sqrt[5]{t^6 \frac{4^{10}}{(t-1)^{10}}} = t^{6/5} \frac{4^2}{(t-1)^2} = t^{6/5} \frac{16}{(t-1)^2}$.

Step 5: Substitute into the integral. $I = \int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} dx = \int \frac{1}{t^{6/5} \frac{16}{(t-1)^2}} \cdot \frac{-4}{(t-1)^2} dt = \int \frac{-4(t-1)^2}{16 t^{6/5} (t-1)^2} dt = \int \frac{-1}{4 t^{6/5}} dt$

Step 6: Evaluate the integral. $I = -\frac{1}{4} \int t^{-6/5} dt = -\frac{1}{4} \cdot \frac{t^{-6/5 + 1}}{-6/5 + 1} + C = -\frac{1}{4} \cdot \frac{t^{-1/5}}{-1/5} + C = \frac{5}{4} t^{-1/5} + C = \frac{5}{4} \left(\frac{x+3}{x-1}\right)^{-1/5} + C = \frac{5}{4} \left(\frac{x-1}{x+3}\right)^{1/5} + C$

Step 7: Compare with the given form. We have $I = A\left(\frac{\alpha x - 1}{\beta x + 3}\right)^B + C$. Comparing with our result, $I = \frac{5}{4} \left(\frac{x-1}{x+3}\right)^{1/5} + C$, we get $A = \frac{5}{4}$, $B = \frac{1}{5}$, $\alpha = 1$, and $\beta = 1$.

Step 8: Calculate $\alpha + \beta + 20AB$. $\alpha + \beta + 20AB = 1 + 1 + 20 \left(\frac{5}{4}\right) \left(\frac{1}{5}\right) = 2 + 20 \cdot \frac{1}{4} = 2 + 5 = 7$.

Common Mistakes & Tips

• Be careful with the chain rule when performing substitutions.
• Double-check algebraic manipulations to avoid errors.
• Remember to substitute back to the original variable after integration.

Summary

We solved the indefinite integral by using the substitution method. By substituting $t = \frac{x+3}{x-1}$, we simplified the integral and evaluated it. Then, by comparing the result with the given form, we found the values of $A$, $B$, $\alpha$, and $\beta$. Finally, we calculated the value of $\alpha + \beta + 20AB$, which is 7.

Final Answer

The final answer is \boxed{7}.