Đặt $\ln x = u \Rightarrow \dfrac{1}{x}dx = du; \dfrac{1}{x^3}dx = dv \Rightarrow -\dfrac{1}{2x^2} = v$$I = -\dfrac{1}{2x^2}\ln x \bigg |_1^2 +\dfrac{1}{2}\int \dfrac{1}{x^3}dx= (-\dfrac{1}{2x^2}\ln x -\dfrac{1}{4x^2}) \bigg |_1^2$
Đặt $\ln x = u \Rightarrow \dfrac{1}{x}dx = du; \dfrac{1}{x^3}dx = dv \Rightarrow -\dfrac{1}{2x^2} = v$$I = -\dfrac{1}{2x^2}\ln x +\dfrac{1}{2}\int \dfrac{1}{x^3}dx= (-\dfrac{1}{2x^2}\ln x -\dfrac{1}{4x^2}) \bigg |_1^2$
Đặt $\ln x = u \Rightarrow \dfrac{1}{x}dx = du; \dfrac{1}{x^3}dx = dv \Rightarrow -\dfrac{1}{2x^2} = v$$I = -\dfrac{1}{2x^2}\ln x
\bigg |_1^2 +\dfrac{1}{2}\int \dfrac{1}{x^3}dx= (-\dfrac{1}{2x^2}\ln x -\dfrac{1}{4x^2}) \bigg |_1^2$