Key Concepts and Formulas

• Integration by Parts: $\int u \, dv = uv - \int v \, du$
• Substitution Method: $\int f(g(x))g'(x) \, dx = \int f(u) \, du$, where $u = g(x)$ and $du = g'(x) \, dx$

Step-by-Step Solution

Step 1: Identify the given integral and the desired form. We are given the integral $\int x^5 e^{-4x^3} \, dx$ and we want to find $f(x)$ such that $\int x^5 e^{-4x^3} \, dx = \frac{1}{48}e^{-4x^3} f(x) + C$.

Step 2: Perform a substitution to simplify the integral. Let $t = x^3$. Then, $dt = 3x^2 \, dx$, which implies $x^2 \, dx = \frac{1}{3} \, dt$. We can rewrite the integral as follows: $\int x^5 e^{-4x^3} \, dx = \int x^3 \cdot x^2 e^{-4x^3} \, dx = \int t e^{-4t} \frac{1}{3} \, dt = \frac{1}{3} \int t e^{-4t} \, dt$ This substitution simplifies the integral and makes it easier to apply integration by parts.

Step 3: Apply integration by parts. We will use integration by parts on $\int t e^{-4t} \, dt$. Let $u = t$ and $dv = e^{-4t} \, dt$. Then, $du = dt$ and $v = \int e^{-4t} \, dt = -\frac{1}{4}e^{-4t}$. Applying the integration by parts formula, we get: $\int t e^{-4t} \, dt = t \left(-\frac{1}{4}e^{-4t}\right) - \int \left(-\frac{1}{4}e^{-4t}\right) \, dt = -\frac{1}{4}te^{-4t} + \frac{1}{4} \int e^{-4t} \, dt$ $= -\frac{1}{4}te^{-4t} + \frac{1}{4} \left(-\frac{1}{4}e^{-4t}\right) + C_1 = -\frac{1}{4}te^{-4t} - \frac{1}{16}e^{-4t} + C_1$

Step 4: Substitute back to express the result in terms of $x$. Now we substitute $t = x^3$ back into the expression: $\frac{1}{3} \int t e^{-4t} \, dt = \frac{1}{3} \left(-\frac{1}{4}x^3e^{-4x^3} - \frac{1}{16}e^{-4x^3}\right) + C = -\frac{1}{12}x^3e^{-4x^3} - \frac{1}{48}e^{-4x^3} + C$

Step 5: Factor out the common term and compare with the given form. We can factor out $-\frac{1}{48}e^{-4x^3}$ from the expression: $-\frac{1}{12}x^3e^{-4x^3} - \frac{1}{48}e^{-4x^3} + C = -\frac{1}{48}e^{-4x^3}(4x^3 + 1) + C$ Since we are given that the integral is equal to $\frac{1}{48}e^{-4x^3} f(x) + C$, we have: $\frac{1}{48}e^{-4x^3} f(x) + C = -\frac{1}{48}e^{-4x^3}(4x^3 + 1) + C$ Therefore, $f(x) = -(4x^3 + 1) = -4x^3 - 1$.

Step 6: Verify the answer We found $f(x) = -4x^3 - 1$. Thus, $\frac{1}{48}e^{-4x^3} f(x) + C = \frac{1}{48}e^{-4x^3} (-4x^3 - 1) + C = -\frac{1}{48}e^{-4x^3} (4x^3 + 1) + C = -\frac{1}{12}x^3e^{-4x^3} - \frac{1}{48}e^{-4x^3} + C$ This matches our integrated result.

Common Mistakes & Tips

• Remember the constant of integration, $C$, when evaluating indefinite integrals.
• Be careful with signs, especially when applying integration by parts.
• When using substitution, make sure to substitute back to the original variable at the end.

Summary

We used substitution and integration by parts to evaluate the given integral. By setting $t = x^3$, we simplified the integral to $\frac{1}{3} \int t e^{-4t} \, dt$. Applying integration by parts, we obtained $-\frac{1}{12}x^3e^{-4x^3} - \frac{1}{48}e^{-4x^3} + C$. Finally, we compared this result with the given form $\frac{1}{48}e^{-4x^3} f(x) + C$ to find that $f(x) = -4x^3 - 1$.

Final Answer

The final answer is \boxed{-4x^3 - 1}, which corresponds to option (D).