Answer
$$\int^e_1 x^3\ln xdx=\frac{3e^4+1}{16}$$
Work Step by Step
$$A=\int^e_1 x^3\ln xdx$$
Set $u=\ln x$ and $dv=x^3dx$
Then we would have $du=\frac{dx}{x}$ and $v=\frac{x^4}{4}$
Using the formula $\int^b_a udv= uv]^b_a-\int^b_a vdu$:
$$A=\frac{x^4\ln x}{4}\Big]^e_1-\int^e_1\Big(\frac{x^4}{4}\times\frac{dx}{x}\Big)$$ $$A=\frac{1}{4}(e^4\ln e-1^4\ln1)-\frac{1}{4}\int^e_1x^3dx$$ $$A=\frac{1}{4}(e^4-0)-\frac{1}{4}\Big(\frac{x^4}{4}\Big)\Big]^e_1$$ $$A=\frac{e^4}{4}-\frac{1}{16}(e^4-1^4)$$ $$A=\frac{4e^4-(e^4-1)}{16}$$ $$A=\frac{3e^4+1}{16}$$
