Tính tích phân: $\int_{1}^{e^2}\frac{x+\ln x-1}{(x\ln x+2)^2}\text{d}x$

Tính tích phân: $\displaystyle \int \limits_{1}^{e^2}\dfrac{x+\ln x-1}{(x\ln x+2)^2}\text{d}x$

Hướng dẫn

Ta có $(x\ln x+2)'=\ln x+1$
$I=\displaystyle \int \limits _{ 1 }^{ e^2}\dfrac{ x+\ln x-1 }{(x\ln x+2)^2} dx \\ =\displaystyle \int \limits_{1}^{e^2}\dfrac { x-2 }{(x\ln x+2)^2} dx +\displaystyle \int \limits_{1}^{e^2}\dfrac {\ln x+1}{(x\ln x+2)^2}dx \\ = \displaystyle \int \limits_{1}^{e^2}\dfrac {\frac {1}{x} -\frac {2}{x^2}}{(\ln x+\frac {2}{x})^2} dx +\displaystyle \int \limits_{1}^{e^2}\dfrac {\ln x+1}{(x\ln x+2)^2}dx =I_1 +I_2$

Tính $I_1$. Đặt $t_1=\ln x+\dfrac {2}{x} \Rightarrow I_1= ... =-\dfrac{e^2}{2e^2+2}+\dfrac{1}{2}$

Tính $I_2$. Đặt $t_2=x\ln x+2 \Rightarrow I_2= ... =-\dfrac{1}{2e^2+2}+\dfrac{1}{2}$

Vậy $I=\dfrac{1}{2}$