Nguyên hàm bài tập phần 5

Quy tắc đồng nhất thức
\( \frac{ax^3 + bx^2 + cx + d}{(x - \alpha)(x - \beta)(\gamma x^2 + \delta x + e)} = \frac{A}{x - \alpha} + \frac{B}{x - \beta} + \frac{Cx + D}{\gamma x^2 + \delta x + \varepsilon} \)

 

Ví dụ:
\( \int \frac{1}{(x - 1)(2x - 2)(x - 3)} \, dx \)  

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 \( (******) \)

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\( \int \frac{3x + 2}{x^2 - 5x + 6} \, dx \)  

\( \frac{3x + 2}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3} \implies  
\begin{cases} 
A = -8 \\ 
B = 11 
\end{cases} \)

\( \int \frac{3x + 2}{x^2 - 5x + 6} \, dx = 11 \int \frac{1}{x - 3} \, dx - 8 \int \frac{1}{x - 2} \, dx \)  

\( = 11 \ln|x - 3| - 8 \ln|x - 2| + C. \)  

 Làm thêm!   
\( \int \frac{5x + 1}{x^2 - 9x + 20} \, dx \)

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3. Tích phân dạng:  
\( \int \frac{\alpha x + \beta}{ax^2 + bx + c} \, dx \) với \( ax^2 + bx + c = 0 \) vô nghiệm.  

 

Ví dụ  
a) \( \int \frac{1}{1 + x^2} \, dx \)  

Đặt \( x = \tan t \implies dx = (1 + \tan^2 t) \, dt \)  

\( \int \frac{1}{1 + x^2} \, dx = \int \frac{1}{1 + \tan^2 t} (1 + \tan^2 t) \, dt = \int dt \)  

\( = t + C = \arctan x + C. \)

Nhớ:  
\( \int \frac{1}{1 + x^2} \, dx = \arctan x + C \)
\(\int \frac{1}{1 + u^2} \, du = \arctan u + C \)
\(\int \frac{u'}{1 + u^2} \, dx = \arctan u + C \)

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b) \( \int \frac{1}{x^2 + 4} \, dx \)

\( \int \frac{1}{x^2 + 4} \, dx = \frac{1}{4} \int \frac{1}{1 + \left( \frac{x}{2} \right)^2} \, dx = \frac{1}{2} \int \frac{\frac{1}{2}}{1 + \left( \frac{x}{2} \right)^2} \, dx \)

\( = \frac{1}{2} \arctan \left( \frac{x}{2} \right) + C. \)

c) \( \int \frac{1}{(x + 1)^2 + 9} \, dx \)
d) \( \int \frac{1}{x^2 - 4x + 6} \, dx \)
e) \( \int \frac{3x + 1}{x^2 + x + 1} \, dx \)

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\( \int \frac{x^2}{(x + 1)^{10}} \, dx \)  

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  

\( \int \frac{(x + 2)^3}{(x - 1)^7} \, dx \)

Đặt \( t = x - 1 \).

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