1. **State the problem:** We need to evaluate the definite integral $$\int_0^1 7x \cdot 7^{x^2} \, dx$$. 2. **Rewrite the integral:** Notice that $$7^{x^2} = e^{x^2 \ln 7}$$, so the integral becomes $$\int_0^1 7x e^{x^2 \ln 7} \, dx$$. 3. **Use substitution:** Let $$u = x^2$$, then $$du = 2x \, dx$$ or $$x \, dx = \frac{du}{2}$$. 4. **Rewrite the integral in terms of $$u$$:** $$\int_0^1 7x e^{x^2 \ln 7} \, dx = 7 \int_0^1 x e^{u \ln 7} \, dx$$ Using substitution, $$= 7 \int_0^1 e^{u \ln 7} \cdot x \, dx = 7 \int_0^1 e^{u \ln 7} \cdot \frac{du}{2} = \frac{7}{2} \int_0^1 e^{u \ln 7} \, du$$ 5. **Evaluate the integral:** $$\int_0^1 e^{u \ln 7} \, du = \int_0^1 7^u \, du = \left[ \frac{7^u}{\ln 7} \right]_0^1 = \frac{7^1 - 7^0}{\ln 7} = \frac{7 - 1}{\ln 7} = \frac{6}{\ln 7}$$ 6. **Combine results:** $$\frac{7}{2} \times \frac{6}{\ln 7} = \frac{42}{2 \ln 7} = \frac{21}{\ln 7}$$ **Final answer:** $$\boxed{\frac{21}{\ln 7}}$$