Let's evaluate the two integrals step-by-step:

(a) $\int xe^{3x} \, dx$

1. Use Integration by Parts Formula: $\int u \, dv = uv - \int v \, du$ Here, let:

   • $u = x$ (so $du = dx$)
   • $dv = e^{3x} dx$ (so $v = \frac{1}{3}e^{3x}$)

2. Apply the Formula: $\int xe^{3x} dx = u \cdot v - \int v \cdot du$ Substituting: $\int xe^{3x} dx = x \cdot \frac{1}{3}e^{3x} - \int \frac{1}{3}e^{3x} dx$

3. Simplify: $\int xe^{3x} dx = \frac{x}{3}e^{3x} - \frac{1}{3} \int e^{3x} dx$ The remaining integral is: $\int e^{3x} dx = \frac{1}{3}e^{3x}$ Substituting back: $\int xe^{3x} dx = \frac{x}{3}e^{3x} - \frac{1}{3} \cdot \frac{1}{3}e^{3x}$ $\int xe^{3x} dx = \frac{x}{3}e^{3x} - \frac{1}{9}e^{3x}$

4. Factorize: $\int xe^{3x} dx = e^{3x} \left(\frac{x}{3} - \frac{1}{9}\right)$ $\int xe^{3x} dx = e^{3x} \left(\frac{3x - 1}{9}\right)$ Final Answer: $\int xe^{3x} dx = \frac{e^{3x}(3x - 1)}{9} + C$

(b) $\int_{1}^{e^4} \frac{\ln x}{x^7} dx$

1. Substitute $u = \ln x$:

   • Then $du = \frac{1}{x} dx$
   • Also, $\ln x = u \implies x = e^u$.

   Substituting limits:

   • When $x = 1$, $u = \ln(1) = 0$
   • When $x = e^4$, $u = \ln(e^4) = 4$

2. Rewrite the Integral: $\int_{1}^{e^4} \frac{\ln x}{x^7} dx = \int_{0}^{4} \frac{u}{e^{7u}} du$

3. Simplify: Factorize: $\int_{0}^{4} \frac{u}{e^{7u}} du = \int_{0}^{4} u \cdot e^{-7u} du$

4. Use Integration by Parts: Let:

   • $v = u$ (so $dv = du$)
   • $dw = e^{-7u} du$ (so $w = -\frac{1}{7}e^{-7u}$)

   Applying the formula: $\int u e^{-7u} du = uv - \int v \, dw$ Substitute: $\int u e^{-7u} du = \left(-\frac{u}{7}e^{-7u}\right) - \int -\frac{1}{7}e^{-7u} du$ $\int u e^{-7u} du = -\frac{u}{7}e^{-7u} + \frac{1}{7} \int e^{-7u} du$ Solve the remaining integral: $\int e^{-7u} du = -\frac{1}{7}e^{-7u}$ Substituting back: $\int u e^{-7u} du = -\frac{u}{7}e^{-7u} - \frac{1}{49}e^{-7u}$ $\int u e^{-7u} du = -e^{-7u} \left(\frac{u}{7} + \frac{1}{49}\right)$

5. Apply the Limits ($u = 0$ to $u = 4$): Substitute $u = 4$ and $u = 0$: $\int_{0}^{4} u e^{-7u} du = -e^{-7(4)} \left(\frac{4}{7} + \frac{1}{49}\right) + e^{-7(0)} \left(\frac{0}{7} + \frac{1}{49}\right)$ Simplify: $\int_{0}^{4} u e^{-7u} du = -e^{-28} \left(\frac{4}{7} + \frac{1}{49}\right) + \frac{1}{49}$

Final Answer for (b) involves simplifying this expression. If you'd like me to compute this explicitly, let me know!

Do you want further explanation or simplifications?
Here are 5 related questions for practice:

1. How would you evaluate $\int e^{kx} dx$ for any constant $k$?
2. How does substitution help simplify integrals involving logarithms?
3. How do you recognize when to use integration by parts?
4. Can you solve $\int_{1}^{e} \frac{1}{x} dx$ and relate it to logarithmic rules?
5. How do definite integrals differ from indefinite integrals when using substitution?

Tip: Always check the bounds of integration when using substitution in definite integrals!