1. **State the problem:** We need to find the integral $$\int \frac{3k^3 + 2k}{8 + k^2} \, dk.$$\n\n2. **Rewrite the integral:** Notice the numerator is a polynomial of degree 3 and the denominator is quadratic. We can try to simplify by polynomial division or splitting the integral.\n\n3. **Split the numerator:** Write $$3k^3 + 2k = 3k^3 + 0k^2 + 2k + 0.$$\n\n4. **Divide numerator by denominator:** Divide $$3k^3 + 2k$$ by $$8 + k^2$$.\n\n- Divide leading terms: $$\frac{3k^3}{k^2} = 3k.$$\n- Multiply divisor by 3k: $$3k(8 + k^2) = 24k + 3k^3.$$\n- Subtract: $$(3k^3 + 2k) - (3k^3 + 24k) = -22k.$$\n\n5. **Rewrite integral:** $$\int \frac{3k^3 + 2k}{8 + k^2} \, dk = \int 3k \, dk + \int \frac{-22k}{8 + k^2} \, dk.$$\n\n6. **Integrate first term:** $$\int 3k \, dk = \frac{3k^2}{2} + C_1.$$\n\n7. **Integrate second term:** Use substitution for $$\int \frac{-22k}{8 + k^2} \, dk.$$\n\n- Let $$u = 8 + k^2,$$ then $$du = 2k \, dk,$$ so $$k \, dk = \frac{du}{2}.$$\n- Rewrite integral: $$\int \frac{-22k}{8 + k^2} \, dk = -22 \int \frac{k}{u} \, dk = -22 \int \frac{1}{u} \cdot k \, dk = -22 \int \frac{1}{u} \cdot \frac{du}{2} = -11 \int \frac{1}{u} \, du.$$\n\n- Integrate: $$-11 \ln|u| + C_2 = -11 \ln|8 + k^2| + C_2.$$\n\n8. **Combine results:** $$\int \frac{3k^3 + 2k}{8 + k^2} \, dk = \frac{3k^2}{2} - 11 \ln|8 + k^2| + C,$$ where $$C = C_1 + C_2$$ is the constant of integration.