Tích phân 3

Tính tích phân $ I=\displaystyle\int_{1}^{2}\dfrac{3-4\ln^2 x}{4x^{2}\sqrt{1+\ln x}}dx$

 Đặt $y=\sqrt{1+\ln x}$, ta có $\ln x=y^2-1$, do đó $\dfrac{1}{y}=y'2x$ và\[\begin{align*}\frac{3-4\ln^{2} x}{4x^{2}\sqrt{1+\ln  x}}&=\frac{3-4\left(y^2-1\right)^2}{4x^2y}\\&=\frac{6y^2-1-\left(4y^4-2y^2\right)}{4x^2y}\\&=\frac{\frac{6y^2-1}{y}-\left(4y^3-2y\right)}{4x^2}\\&=\frac{(6y^2-1)y'2x-\left(2y^3-y\right)(2x)'}{(2x)^2}\\&=\frac{(2y^3-y)'2x-\left(2y^3-y\right)(2x)'}{(2x)^2}\\&=\left(\frac{2y^3-y}{2x}\right)'\\ &=\left(\dfrac{2\left(\sqrt{1+\ln x}\right)^3-\sqrt{1+\ln x}}{2x}\right)'\end{align*}\]Vậy $I=\left(\dfrac{2\left(\sqrt{1+\ln x}\right)^3-\sqrt{1+\ln x}}{2x}\right)\Bigg|_1^2=\dfrac{2\left(\sqrt{1+\ln 2}\right)^3-\sqrt{1+\ln 2}-2}{4}$