$I = \displaystyle \int{\dfrac{dx}{(x-2)^2.x}}$
Ta có: $\dfrac{1}{(x-2)^2.x} = \dfrac{A}{(x-2)^2}+\dfrac{B}{x-2}+\dfrac{C}{x} \\
\\ = \dfrac{Ax+B(x-2)+C(x-2)^2}{x.(x-2)^2} \\
= \dfrac{Ax+Bx-2B+C(x^2-4x+4)}{x.(x-2)^2} \\
\\
= \dfrac{Ax+Bx-2B+Cx^2-4Cx+4C}{x.(x-2)^2} \\
\\
= \dfrac{Cx^2+(A+B-4C)x-2B+4C}{x.(x-2)^2}$
Đồng nhất hệ số, ta có hệ phương trình:
$
\left\{\begin{matrix}
C=0 \\
A+B-4C=0 \\
-2B+4C=1
\end{matrix}\right. \\
\Leftrightarrow \left\{\begin{matrix}
A = \dfrac{1}{2} \\
\\
B= - \dfrac{1}{2} \\
\\
C=0
\end{matrix}\right. \\
$
Do đó $\dfrac{1}{(x-2)^2.x} = \dfrac{1}{2(x-2)^2} - \dfrac{1}{2(x-2)}$
$\displaystyle I = \int{\dfrac{dx}{(x-2)^2.x}} = \int{\left [ \dfrac{1}{2(x-2)^2} - \dfrac{1}{2(x-2)} \right ]}dx \\
\displaystyle = \dfrac{1}{2} \int{\dfrac{dx}{(x-2)^2}} - \dfrac{1}{2} \int{\dfrac{dx}{x-2}} \\
= \dfrac{1}{2} I_1 - \dfrac{1}{2} I_2$
* Tính $\displaystyle I_1 = \int{\dfrac{dx}{(x-2)^2}}$
$\displaystyle I_1 = \int{\dfrac{dx}{(x-2)^2}} = \int{(x-2)^{-2}}dx = \dfrac{(x-2)^{-1}}{-1} = - \dfrac{1}{x-2} + C_1$
* Tính $\displaystyle I_2 = \int{\dfrac{dx}{x-2}}$
$\displaystyle I_2 = \int{\dfrac{dx}{x-2}} = \ln{|x-2|}+C_2$
Vậy $I = - \dfrac{1}{2(x-2)} - \dfrac{\ln{|x-2|}}{2}+C$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
