Tìm \( A, B, C \) sao cho
\( \frac{1}{(x - 1)(x - 2)(x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3} \)
\( = \frac{A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)}{(x - 1)(x - 2)(x - 3)} \)
\( = \frac{(A + B + C)x^2 + (-5A - 4B - 3C)x + 6A + 3B + 2C}{(x - 1)(x - 2)(x - 3)} \)
\( \Leftrightarrow
\begin{cases}
A + B + C = 0 \\
-5A - 4B - 3C = 0 \\
6A + 3B + 2C = 1
\end{cases}
\Leftrightarrow
\begin{cases}
A = \frac{1}{2} \\
B = -1 \\
C = \frac{1}{2}
\end{cases} \)
Cách 2: Dùng giới hạn \( (******) \)
page25
\( \frac{3x + 2}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3} \implies
\begin{cases}
A = -8 \\
B = 11
\end{cases} \)
\( \int \frac{3x + 2}{x^2 - 5x + 6} \, dx = 11 \int \frac{1}{x - 3} \, dx - 8 \int \frac{1}{x - 2} \, dx \)
\( = 11 \ln|x - 3| - 8 \ln|x - 2| + C. \)
page26
Đặt \( x = \tan t \implies dx = (1 + \tan^2 t) \, dt \)
\( \int \frac{1}{1 + x^2} \, dx = \int \frac{1}{1 + \tan^2 t} (1 + \tan^2 t) \, dt = \int dt \)
\( = t + C = \arctan x + C. \)
page27
\( \int \frac{1}{x^2 + 4} \, dx = \frac{1}{4} \int \frac{1}{1 + \left( \frac{x}{2} \right)^2} \, dx = \frac{1}{2} \int \frac{\frac{1}{2}}{1 + \left( \frac{x}{2} \right)^2} \, dx \)
\( = \frac{1}{2} \arctan \left( \frac{x}{2} \right) + C. \)
page28
Đặt \( t = x + 1 \implies x = t - 1 \implies dx = dt \)
\( \int \frac{x^2}{(x + 1)^{10}} \, dx = \int \frac{(t - 1)^2}{t^{10}} \, dt \)
\( = \int \left( \frac{1}{t^8} - \frac{2}{t^9} + \frac{1}{t^{10}} \right) \, dt = -\frac{8}{t^7} + \frac{18}{t^8} - \frac{10}{t^9} + C \)
\( = -\frac{8}{(x + 1)^7} + \frac{18}{(x + 1)^8} - \frac{10}{(x + 1)^9} + C. \)
Làm thêm
Đặt \( t = x - 1 \).
page29