Lời giải:

 \( I = \int_1^e \frac{\ln x - 1}{x^2 \left[ \left( \frac{\ln x}{x} \right)^2 - 1 \right]} \, dx \)

Đặt \( t = \frac{\ln x}{x} \quad \Rightarrow \quad dt = \frac{(1 - \ln x)}{x^2} \, dx \)

 \(\begin{cases}
x = 1 \Rightarrow t = 0   \\
x = e  \Rightarrow t = \frac{1}{e}
\end{cases}\)

\( I = - \int_0^{\frac{1}{2}} \frac{1}{t^2 - 1} \, dt \)

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Chú ý !  \( I = \int_0^{\frac{\pi}{2}} \frac{1}{3 \sin^2 x + \cos^2 x} \, dx \)

Không thế đặt: \( t = \tan x \), hoặc \( t = \cot x \)

\( I = \int_0^{\frac{\pi}{4}} \frac{1}{3\sin^2 x  + \cos^2 x } \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{3\sin^2 x  + \cos^2 x } \, dx \)

Đặt  \( t = \tan x \):  
\( \Rightarrow I_1 = \int_0^{\frac{\pi}{4}} \frac{1}{\cos^2 x \left[ 3 \tan^2 x + 1 \right]} \, dx \)

Đặt \( t = \cot x \):  
\(\Rightarrow  I_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{\sin^2 x \left[ 3 + \cot^2 x \right]} \, dx \)

\( \Leftrightarrow I_1 + I2 =  \frac{\pi \sqrt{3}}{6} \)

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Lời giải

 \( I = \int_0^\pi \sqrt{(\cos x + \sin x)^2} \, dx = \int_0^\pi |\cos x + \sin x| \, dx \)

Gợi ý:  \( \cos x + \sin x = 0 \Leftrightarrow \tan x = -1 \Leftrightarrow x = \frac{3\pi}{4} \) (trong \( (0, \pi) \))

\( I = \int_0^{\frac{3\pi}{4}} (\cos x + \sin x) \, dx - \int_{\frac{3\pi}{4}}^\pi (\cos x + \sin x) \, dx \)

  \( = (\sin x - \cos x ) \bigg|_0^{\frac{3\pi}{4}} - ( \sin x - \cos x ) \bigg|_{\frac{3\pi}{4}}^\pi \)

  \(  = (\sqrt{2} + 1) - (1-\sqrt{2}) = 2\sqrt{2} \)

Bấm máy tính: (Chuyển máy tính về chế độ radian Shift → Mode → 4)

• \( \int_0^\pi \sqrt{1 + \sin 2x} \, dx = 2,828427 \)

• \( 2 \sqrt{2} =  2.2^\frac{1}{2} = 2.828427 \)

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Lời giải

* \( 2^x - 2^{-x} > 0 \iff 2^x > 2^{-x} \iff x >-x \iff x > 0 \)

\( \int_{-1}^1 |2^x - 2^{-x}| \, dx = \int_{-1}^0 |2^x - 2^{-x}| \, dx + \int_0^1 |2^x - 2^{-x}| \, dx \)

\(  = \int_{-1}^0 (2^{-x} - 2^x) \, dx + \int_0^1 (2^x - 2^{-x}) \, dx \)

\( = \left( -\frac{2^{-x}}{\ln 2} - \frac{2^x}{\ln 2} \right) \bigg|_{-1}^0 + \left( \frac{2^x}{\ln 2} + \frac{2^{-x}}{\ln 2} \right) \bigg|_0^1 \)

\(  = \left( -\frac{2}{\ln 2} + \frac{2}{\ln 2} + \frac{1}{2\ln 2} \right) + \left( \frac{2}{\ln2} + \frac{1}{2\ln 2} \ -\frac{2}{\ln 2} \right) \)

\(  = \frac{1}{\ln 2}\)  (Bấm máy= 1.442695)

* Bấm:  \( \int_{-1}^1 \sqrt{(2^x - 2^{-x})^2} \, dx = 1.442695 \)

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Lời giải

\( I = \int_0^3 \sqrt{x(x^2 - 2x + 1)} \, dx = \int_0^3 \sqrt{x(x - 1)^2} \, dx \)

\(= \int_0^3 |x - 1| \sqrt{x}\, dx\)

\(= \int_0^1 (1 - x) \sqrt{x} \, dx + \int_1^3 (x - 1) \sqrt{x} \, dx\)

\(= \int_0^1 (x^{\frac{1}{2}} - x^{\frac{3}{2}}) \, dx + \int_1^3 (x^{\frac{3}{2}} - x^{\frac{1}{2}}) \, dx\)

\(= \left( \frac{2}{3} x \sqrt{x}- \frac{2}{5} x^2 \sqrt{x}\right)\bigg|_0^1 + \left( \frac{2}{5} x^2 \sqrt{x} - \frac{2}{3} x \sqrt{x} \right)\bigg|_1^3\)

\(= \frac{8 + 24\sqrt{3}}{15}\)

* Bấm 
•  \( \int_0^3 \sqrt{x^3 - 2x^2 + x} \, dx = 3.304614\)

•  \( \frac{8 + 24.3^{\frac{1}{2}}}{15} = 3.304614 \)

\(\Rightarrow\) Vậy chọn đáp án \(\boxed{\text{B}} \)  

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