Đặt \(u=\ln^2x\rightarrow du=2\ln x\frac{dx}{x},dv=\int\limits x^3dx\rightarrow v=\frac{1}{4}x^4\)
Do đó : \(I=\frac{1}{4}x^4.\ln^2x|^e_1-\frac{1}{4}\int\limits^e_12\ln x.\frac{x^4}{x}dx=\frac{e^4}{4}-\frac{1}{2}\int\limits^e_1x^3\ln sdx=\frac{e^4}{4}-\frac{1}{2}J\left(1\right)\)
Tính \(J=\int\limits^e_1x^3\ln xdx\)
Đặt \(u_1=\ln x\rightarrow du_1=\frac{dx}{x},dv_1=\int x^3dx\rightarrow v_1=\frac{1}{4}x^4\)
Do đó :
\(J=\frac{1}{4}x^4\ln x|^e_1-\frac{1}{4}\int\limits^e_1x^3dx=\frac{e^4}{4}-\frac{1}{16}x^2|^e_1=\frac{3e^4+1}{16}\)
Thay vào (1) ta có :
\(I=\frac{e^4}{4}-\frac{1}{2}\left(\frac{3e^4+1}{16}\right)=\frac{5e^4-1}{32}\)
