We are tasked with solving the integral:

$\int x^3 e^{4x} \, dx$

To solve this, we will use integration by parts, which is based on the formula:

$\int u \, dv = uv - \int v \, du$

Step 1: Choose $u$ and $dv$

Let:

• $u = x^3$ (we choose this because its derivative simplifies the expression)
• $dv = e^{4x} dx$

Step 2: Differentiate and Integrate

Now, differentiate $u$ and integrate $dv$:

• $du = 3x^2 \, dx$
• $v = \frac{e^{4x}}{4}$ (the integral of $e^{4x}$ is $\frac{e^{4x}}{4}$)

Step 3: Apply the Integration by Parts Formula

Now, we apply the formula $\int u \, dv = uv - \int v \, du$:

$\int x^3 e^{4x} \, dx = \left( x^3 \cdot \frac{e^{4x}}{4} \right) - \int \left( \frac{e^{4x}}{4} \cdot 3x^2 \right) dx$ $= \frac{x^3 e^{4x}}{4} - \frac{3}{4} \int x^2 e^{4x} \, dx$

Step 4: Repeat Integration by Parts for $\int x^2 e^{4x} \, dx$

Next, we need to integrate $\int x^2 e^{4x} \, dx$. We use integration by parts again:

• Let $u = x^2$ and $dv = e^{4x} dx$
• $du = 2x \, dx$ and $v = \frac{e^{4x}}{4}$

Applying the formula again:

$\int x^2 e^{4x} \, dx = \left( x^2 \cdot \frac{e^{4x}}{4} \right) - \int \left( \frac{e^{4x}}{4} \cdot 2x \right) dx$ $= \frac{x^2 e^{4x}}{4} - \frac{2}{4} \int x e^{4x} \, dx$ $= \frac{x^2 e^{4x}}{4} - \frac{1}{2} \int x e^{4x} \, dx$

Step 5: Repeat Integration by Parts for $\int x e^{4x} \, dx$

Now, we handle $\int x e^{4x} \, dx$. Again using integration by parts:

• Let $u = x$ and $dv = e^{4x} dx$
• $du = dx$ and $v = \frac{e^{4x}}{4}$

$\int x e^{4x} \, dx = \left( x \cdot \frac{e^{4x}}{4} \right) - \int \left( \frac{e^{4x}}{4} \cdot 1 \right) dx$ $= \frac{x e^{4x}}{4} - \frac{1}{4} \int e^{4x} \, dx$ $= \frac{x e^{4x}}{4} - \frac{1}{4} \cdot \frac{e^{4x}}{4}$ $= \frac{x e^{4x}}{4} - \frac{e^{4x}}{16}$

Step 6: Substitute Back into the Original Expression

Now, substitute this result back into the previous equations: $\int x^2 e^{4x} \, dx = \frac{x^2 e^{4x}}{4} - \frac{1}{2} \left( \frac{x e^{4x}}{4} - \frac{e^{4x}}{16} \right)$ $= \frac{x^2 e^{4x}}{4} - \frac{1}{2} \cdot \frac{x e^{4x}}{4} + \frac{1}{32} e^{4x}$ $= \frac{x^2 e^{4x}}{4} - \frac{x e^{4x}}{8} + \frac{e^{4x}}{32}$

Now substitute this into the equation for the integral of $x^3 e^{4x}$: $\int x^3 e^{4x} \, dx = \frac{x^3 e^{4x}}{4} - \frac{3}{4} \left( \frac{x^2 e^{4x}}{4} - \frac{x e^{4x}}{8} + \frac{e^{4x}}{32} \right)$ $= \frac{x^3 e^{4x}}{4} - \frac{3}{16} x^2 e^{4x} + \frac{3}{32} x e^{4x} - \frac{3}{128} e^{4x}$

Step 7: Final Answer

Thus, the integral is:

$\int x^3 e^{4x} \, dx = \frac{x^3 e^{4x}}{4} - \frac{3 x^2 e^{4x}}{16} + \frac{3 x e^{4x}}{32} - \frac{3 e^{4x}}{128} + C$

Where $C$ is the constant of integration.

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Would you like further details or clarifications on any of the steps?

Further Exploration:

1. How does the integration by parts formula generalize to other functions?
2. What happens if we apply a substitution method instead of integration by parts in this case?
3. How can you solve integrals involving powers of $x$ and exponential functions efficiently?
4. What is the role of the constant of integration in the final solution?
5. Can we use this method to solve integrals with higher powers of $x$?

Tip:

For integrals of the form $\int x^n e^{ax} \, dx$, repeated integration by parts is often necessary, and the general form will involve a series of decreasing powers of $x$.