Question

Tính tích phân :

\(\int\limits^e_1\ln^3xdx\)

Answer 1

Đặt \(u=\ln^3x\rightarrow du=3\ln^2x\frac{dx}{x},dv=dx\rightarrow v=x\)

Do đó : \(I=x\ln^3x|^e_1-3\int\limits^3_1\ln^2xdx=e-3J\left(1\right)\)

Tính \(J=\int\limits^e_1\ln^2xdx\)

Đặt \(u_1=\ln^2x\rightarrow du_1=\frac{2\ln x}{x}dx,dv_1=dx\rightarrow v_1=x\)

Do vậy, \(J=x\ln^2x|^e_1-2\int\limits^e_1\ln xdx=e-2\left(x\ln x|^e_1-\int\limits^e_1dx\right)=e-2\left(x\ln x-x\right)|^e_1=e-2\)

Thay vào (1) ta có : \(I=e-3\left(e-2\right)=6-2e\)