Đặt \( u = 2x \Rightarrow du = 2dx \),
\( \int \frac{1}{\cos^2 2x} \, dx = \frac{1}{2} \int \frac{1}{\cos^2 u} \, du = \frac{1}{2} \tan u + C. \)
\( \int \frac{1}{1 + \sin 2x} \, dx = \int \frac{1}{(\cos x + \sin x)^2} \, dx = \int \frac{1}{[\sqrt{2} \cos (x - \frac{\pi}{4})]^2} \, dx \)
\( = \frac{1}{2} \int \frac{1}{\cos^2 (x - \frac{\pi}{4})} \, dx = \frac{1}{2} \tan (x - \frac{\pi}{4}) + C. \)
Tương tự: \( \int \frac{1}{1 - \sin 2x} \, dx \)
page20
Tích phân của hàm số hữu tỷ bậc 2
\( \int \frac{1}{ax^2 + bx + c} \, dx = \frac{1}{a} \int \frac{1}{(x - x_0)^2} \, dx = -\frac{1}{a(x - x_0)} + C \)
Ví dụ:
a) \( \int \frac{1}{x^2 - 6x + 9} \, dx \)
b) \( \int \frac{2x - 6}{x^2 - 6x + 9} \, dx \)
c) \( \int \frac{2x + 1}{x^2 - 6x + 9} \, dx \)
d) \( \int \frac{3x - 2}{x^2 - 6x + 9} \, dx \)
e) \( \int \frac{2x^2 + 4x - 1}{x^2 - 6x + 9} \, dx \)
Làm thêm:
\( \int \frac{3x^2 - 4x + 2}{x^2 - 4x + 4} \, dx \)
page21
Tìm A và B sao cho
\( \frac{1}{x^2 - 4x + 3} = \frac{1}{(x - 1)(x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 3} \)
\( =\frac{(A + B)x - 3A - B}{(x - 1)(x - 3)} \)
\( \Leftrightarrow \) \(\begin{cases}
A + B = 0 \\
-3A - B = 1
\end{cases}
\Leftrightarrow
\begin{cases}
A = -\frac{1}{2} \\
B = \frac{1}{2}
\end{cases}
\)
\( \int \frac{1}{x^2 - 4x + 3} \, dx = -\frac{1}{2} \int \frac{1}{x - 1} \, dx + \frac{1}{2} \int \frac{1}{x - 3} \, dx \)
\(= + \frac{1}{2} \left( \ln|x - 3| - \ln|x - 1| \right) = \frac{1}{2} \ln \left| \frac{x - 3}{x - 1} \right| + C\)
Cách 2: Tính nhẩm
page22
\( \int \frac{1}{x^2 + x - 6} \, dx = \int \frac{1}{(x + 3)(x - 2)} \, dx = \frac{1}{5} \int \left( \frac{1}{x - 2} - \frac{1}{x + 3} \right) \, dx. \)
Cách 1: Đồng nhất thức
\( \frac{2x + 2}{x^2 + x - 6} = \frac{A}{x - 2} + \frac{B}{x + 3} \)
\(
\Leftrightarrow
\begin{cases}
A + B = 2 \\
3A - 2B = 2
\end{cases}
\Leftrightarrow
\begin{cases}
A = \frac{6}{5} \\
B = \frac{4}{5}
\end{cases}
\)
Cách 2:
\( \int \frac{2x + 2}{x^2 + x - 6} \, dx = \int \frac{2x + 1}{x^2 + x - 6} \, dx + \int \frac{1}{(x - 2)(x + 3)} \, dx \)
\( = \ln|x^2 + x - 6| + \frac{1}{5} \ln \left| \frac{x - 2}{x + 3} \right| + C. \)
page23
\( \int \frac{2x^2 + 2x - 1}{x^2 - 5x + 4} \, dx = \int \frac{2(x^2 - 5x + 4) + 12x + 9}{x^2 - 5x + 4} \, dx \)
\( = \int 2 \, dx + 6 \int \frac{2x - 5}{x^2 - 5x + 4} \, dx + 13 \int \left( \frac{1}{x - 4} - \frac{1}{x - 1} \right) \, dx \)
\( = 2x + \ln|x^2 - 5x + 4| + 13 \ln \left| \frac{x - 4}{x - 1} \right| + C. \)
Làm thêm:
\( \int \frac{3x^2 + 2x - 4}{x^2 - 7x + 12} \, dx \)
page24