evaluate the following integral. \n\\(\\int_{...

# Explanation: ## Step1: Apply integration - by - parts formula The integration - by - parts formula is $\int_{a}^{b}u\;dv=uv|_{a}^{b}-\int_{a}^{b}v\;du$. Let $u = \ln(x)$ and $dv=x^{3}dx$. Then $du=\frac{1}{x}dx$ and $v=\frac{1}{4}x^{4}$. ## Step2: Calculate $uv|_{a}^{b}$ and $\int_{a}^{b}v\;du$ $uv|_{1}^{e^{7}}=\frac{1}{4}x^{4}\ln(x)\big|_{1}^{e^{7}}=\frac{1}{4}(e^{7})^{4}\ln(e^{7})-\frac{1}{4}(1)^{4}\ln(1)=\frac{7}{4}e^{28}-0=\frac{7}{4}e^{28}$. $\int_{1}^{e^{7}}v\;du=\int_{1}^{e^{7}}\frac{1}{4}x^{4}\cdot\frac{1}{x}dx=\frac{1}{4}\int_{1}^{e^{7}}x^{3}dx$. ## Step3: Evaluate $\frac{1}{4}\int_{1}^{e^{7}}x^{3}dx$ Using the power - rule for integration $\int x^{n}dx=\frac{1}{n + 1}x^{n+1}+C(n\neq - 1)$, we have $\frac{1}{4}\int_{1}^{e^{7}}x^{3}dx=\frac{1}{4}\cdot\frac{1}{4}x^{4}\big|_{1}^{e^{7}}=\frac{1}{16}((e^{7})^{4}-1^{4})=\frac{1}{16}(e^{28}-1)$. ## Step4: Calculate the original integral $\int_{1}^{e^{7}}x^{3}\ln(x)dx=\frac{7}{4}e^{28}-\frac{1}{16}(e^{28}-1)=\frac{28e^{28}-e^{28}+1}{16}=\frac{27e^{28}+1}{16}$. # Answer: $\frac{27e^{28}+1}{16}$