\(I=\frac{1}{4}\int\limits^e_1\frac{4\ln^2x-1+1}{x\left(1+2\ln x\right)}dx=\frac{1}{4}\int\limits^e_1\frac{\left(2\ln x-1\right)dx}{x}+\frac{1}{4}\int\limits^e_1\frac{dx}{x\cdot\left(1+2\ln x\right)}\)

  \(=\frac{1}{8}\int\limits^e_1\left(2\ln x-1\right)d\left(2\ln x-1\right)+\frac{1}{8}\int\limits^e_1\frac{d\left(2\ln x+1\right)}{\left(1+2\ln x\right)}\)

   \(=\left(\frac{1}{16}\left(2\ln x-1\right)^2\right)|^e_1+\frac{1}{8}\ln\left|\left(1+2\ln x\right)\right||^e_1\)

    \(=\frac{1}{8}\ln3\)

Ta có \(I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{\ln2.\ln\left(2\tan x\right)}{\sin2x.\ln\left(2\tan x\right)}dx=\ln2\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x.\ln\left(2\tan x\right)}+\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x}\)

Tính \(\ln2\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x.\ln\left(2\tan x\right)}=\frac{\ln2}{2}\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{d\left[\ln\left(2\tan x\right)\right]}{\ln2\left(2\tan x\right)}=\frac{\ln2}{2}\left[\ln\left(\ln\left(2\tan x\right)\right)\right]|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{\ln2}{2}.\ln\left(\frac{\ln2\sqrt{3}}{\ln2}\right)\)

Tính \(\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x}=\frac{1}{2}\ln\left(\tan x\right)|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{1}{2}\ln\sqrt{3}\)

Vậy \(I=\frac{\ln2}{2}\ln\left(\frac{\ln2\sqrt{3}}{\ln2}\right)+\frac{1}{2}\ln\sqrt{3}\)

\(I=\int\limits^5_1\left(\frac{x}{\sqrt{x-1}+1}+\frac{\ln x}{\left(x+1\right)^2}\right)dx=\int\limits^5_1\frac{x}{\sqrt{x-1}+1}dx+\int\limits^5_1\frac{\ln x}{\left(x+1\right)^2}dx\)

- Tính \(\int\limits^5_1\frac{x}{\sqrt{x-1}+1}dx\)

Đặt \(t=\sqrt{x-1}\Rightarrow t^2=x-1\Leftrightarrow x=t^2+1\Rightarrow dx=2tdt\)

Đổi cận : Cho x=1 => t=0; x=5=>t=2

\(I_1=\int\limits^2_0\frac{t^2+1}{t+1}.2td=\int\limits^2_0\frac{2t^3+2t}{t+1}dt=\int\limits^2_0\left(2t^2-2t+4-\frac{4}{t+1}\right)dt\)

    \(=\left(\frac{2}{3}t^3-t^2+4t-4\ln\left|x+1\right|\right)|^2_0=\frac{28}{3}-4\ln3\)

\(I_2=\int\limits^5_1\frac{\ln x}{\left(x+1\right)^2}dx\)

Đặt \(\begin{cases}u=\ln x\\dv=\frac{1}{\left(x+1\right)^2}dx\end{cases}\) \(\Rightarrow\begin{cases}du=\frac{1}{x}dx\\v=-\frac{1}{x+1}\end{cases}\)

Ta có \(I_2=-\frac{1}{x+1}\ln x|^5_1+\int\limits^5_1\frac{1}{x\left(x+1\right)}dx=-\frac{1}{6}\ln5+\int\limits^5_1\left(\frac{1}{x}-\frac{1}{x+1}\right)dx\)

\(=-\frac{1}{6}\ln5+\left(\ln\left|x\right|x+1\right)|^5_1=-\frac{1}{6}\ln5+\ln5-\ln6+\ln2=\frac{5}{6}\ln5-\ln3\)

Khi đó \(I=I_1+I_2=\frac{28}{3}+\frac{5}{6}\ln5=5\ln3\)

a) Với x > 0 bất kì và \(h = x - {x_0}\) ta có

\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {{x_0} + h} \right) - f\left( {{x_0}} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {{x_0} + h} \right) - \ln {x_0}}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {1 + \frac{h}{{{x_0}}}} \right)}}{{\frac{h}{{{x_0}}}.{x_0}}} = \mathop {\lim }\limits_{h \to 0} \frac{1}{{{x_0}}}.\mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {1 + \frac{h}{{{x_0}}}} \right)}}{{\frac{h}{{{x_0}}}}} = \frac{1}{{{x_0}}}\end{array}\)

Vậy hàm số \(y = \ln x\) có đạo hàm là hàm số \(y' = \frac{1}{x}\)

b) Ta có \({\log _a}x = \frac{{\ln x}}{{\ln a}}\) nên \(\left( {{{\log }_a}x} \right)' = \left( {\frac{{\ln x}}{{\ln a}}} \right)' = \frac{1}{{x\ln a}}\)

Ta có : \(I=\int\limits^2_12x^3dx+\int\limits^2_1\ln xdx\)

Đặt \(I_1=\int\limits^2_12x^3dx\) và \(I_2=\int\limits^2_1\ln xdx\)

Ta có : 

\(I_1=\frac{1}{2}x^4|^2_1=\frac{15}{2}\)

\(I_2=x.\ln x|^2_1-\int_1xd^2\left(\ln x\right)=2\ln2-x|^2_1=2\ln2-1\)

Vậy \(I=I_1+I_2=\frac{13}{2}+2\ln2\)