Đặt: \( \begin{cases} u = \cos(\ln{x}) \\ dv = dx \end{cases} \quad \Rightarrow \quad \begin{cases} du = -\frac{1}{x} \sin(\ln{x}) \, dx \\ v = x \end{cases} \)

\( \int \cos(\ln{x}) \, dx = x \cos(\ln{x}) + \int \sin(\ln{x}) \, dx. \)

Đặt: \( \begin{cases} u = \sin(\ln{x}) \\ dv = dx \end{cases} \quad \Rightarrow \quad \begin{cases} du = \frac{1}{x} \cos(\ln{x}) \, dx \\ v = x \end{cases} \)

\( \int \cos(\ln{x}) \, dx = x \cos(\ln{x}) + x \sin(\ln{x}) - \int \cos(\ln{x}) \, dx. \)

\( \Rightarrow \int \cos(\ln{x}) \, dx = \frac{1}{2} x (\cos(\ln{x}) + \sin(\ln{x})) + C. \)

\(\int \frac{1 + \cos(2\ln{x})}{2} \, dx\)

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Xét: \( \int (1 + \tan^2{x}) e^x \, dx. \)

Đặt: \( \begin{cases} u = e^x \\ dv = (1 + \tan^2{x}) \, dx \end{cases} \quad \Rightarrow \quad \begin{cases} du = e^x \, dx \\ v = \tan{x} \end{cases} \)

\( \int (1 + \tan^2{x}) e^x \, dx = e^x \tan{x} - \int \tan{x} \cdot e^x \, dx. \)

\( \Rightarrow \int (\tan^2{x} + \tan{x} + 1) e^x \, dx = e^x \tan{x} + C. \)

page101

Xét: \( \int \frac{1}{\ln{x}} \, dx. \)

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

\( \int \frac{1}{\ln{x}} \, dx =  \frac{x}{\ln{x}} + \int \frac{1}{\ln^2{x}} \, dx. \)

\( \Rightarrow \int \left( \frac{1}{\ln{x}} - \frac{1}{\ln^2{x}} \right) dx = \frac{x}{\ln{x}} + C. \)

page102

 \( f'(1) = 0 \Leftrightarrow a + b = 0 \quad \Rightarrow \quad b = -a. \)

\( f(x) = \int f'(x) \, dx = \int \left( ax - \frac{a}{x^2} \right) dx = \frac{a x^2}{2} + \frac{a}{x} + C. \)

\( f(-1) = 4 \Leftrightarrow  \frac{a}{2} - a + C = 4 \Leftrightarrow  -\frac{a}{2} + C = 4. \)

 \( f(1) = 2 \Leftrightarrow \frac{a}{2} + a + C = 2 \quad \Rightarrow \quad \frac{3}{2}a  + C = 2. \)

\( \Leftrightarrow \begin{cases} 2a  = -2 \\   C = 4+ \frac{a}{2} \end{cases} \quad \Leftrightarrow \begin{cases} a=-1 \\   C = \frac{7}{2} \end{cases} \)

\( \Rightarrow f(x) = -\frac{x^2}{2} - \frac{1}{x} + \frac{7}{2} \Rightarrow  \boxed{C} \)

page103

Đặt: \( u = 3x + 1 \quad \Rightarrow \quad du = 3 \, dx. \)

\( \int f(3x + 1) \, dx = \frac{1}{3} \int f(u) \, du = \frac{1}{3} F(u) + C = \frac{1}{3} F(3x + 1) + C \Rightarrow  \boxed{C} \)

Cách 2:

\( \int f(3x + 1) \, dx = \frac{1}{3} \int f(3x + 1) \, d(3x + 1) = \frac{1}{3} F(3x + 1) + C. \)

page104

Biết: \( \int f(u) \, du = 2 u^2 + u + C\)

Đặt: \( u = x^2 \quad \Rightarrow \quad du = 2x \, dx. \)

\( \int f(x^2) \, dx = \int f(u) \cdot \frac{1}{2x} \, du  = \frac{1}{2} \int \frac{1}{\sqrt{u}} f(u) \, du. \)

* \(\int f(x) \, dx = 2x^2 + x + C\)

\(\Rightarrow f(x) = 4x + 1 \Rightarrow f(x^2) = 4x^2 + 1\)

\(\Rightarrow \int f(x^2) \, dx = \int (4x^2 + 1) \, dx = \frac{4x^3}{3} + x + C \Rightarrow  \boxed{C}\)

page105

( \int f(x) \, dx = \frac{x - 2}{x + 1} + C \)

\(\Rightarrow f(x) = \frac{3}{(x + 1)^2} \Rightarrow f(e^x) = \frac{3}{(1 + e^x)^2}\)

\( \int f(e^x) \, dx = 3 \int \frac{1}{(1 + e^x)^2} \, dx = 3 \left[ \int \frac{1 + e^x - e^x}{(1 + e^x)^2} \, dx \right] \)

\( = 3 \left( \int \frac{1}{1 + e^x} \, dx - \int \frac{e^x}{(1 + e^x)^2} \, dx \right) \)

\( = 3 \left( \int \frac{1 + e^x - e^x}{1 + e^x} \, dx + \frac{1}{1 + e^x} + C \right) \)

\( = 3 \left[ x - \ln|1 + e^x| + \frac{1}{1 + e^x} \right] + C \Rightarrow  \boxed{C}\)

page106

Một cách giảng độc đáo:  
Cách ra đề:  
 Xét \( u = \frac{x^2}{1 + x^2} \Rightarrow u' = \frac{2x}{(1 + x^2)^2} \)

Ra để:  \( \int \frac{2x}{(1 + x^2)^2} \cdot \left( \frac{x^2}{1 + x^2} \right)^{1006} dx = 2 \int \frac{x^{2013}}{(1 + x^2)^{1008}} \, dx \)

\( \int \frac{x^{2013}}{(1 + x^2)^{1008}} \, dx = \frac{1}{2} \int \frac{2x}{(1 + x^2)^2} \cdot \left( \frac{x^2}{1 + x^2} \right)^{1006} dx \)

\( = \frac{1}{2} \int u'u^{1006} \, du = \frac{1}{2} \frac{u^{1007}}{1007} + C \)

\( = \frac{1}{2014} \left( \frac{x^2}{1 + x^2} \right)^{1007} + C \)

page107

\( I = \int \frac{\ln x - 1}{x^2 (1 - \left( \frac{\ln x}{x} \right)^2)} \, dx\)

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

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

\( I = \frac{1}{2} \int \left( \frac{1}{t - 1} - \frac{1}{t + 1} \right) \, dt = \frac{1}{2} \ln \left| \frac{t - 1}{t + 1} \right| + C \)

page108