题目
求不定积分：
$$
\int \frac{3x + 6}{(x – 1)^{2} (x^{2} + x + 1)} \mathrm{d} x.
$$
解析
本题需要用到待定系数法计算不定积分，关于该方法的详细介绍，可以参考这篇文章：
令：
$$
\int \frac{3x + 6}{(x – 1)^{2} (x^{2} + x + 1)} \mathrm{d} x =
$$
$$
\int \Bigg[ \frac{Ax + B}{(x – 1)^{2}} + \frac{Cx + D}{x^{2} + x + 1} \Bigg] \mathrm{d} x \Rightarrow
$$
$$
\color{White}{
\blacktriangledown
}
$$
上式中的 $\frac{Ax + B}{(x – 1)^{2}}$ 不易计算积分，因此还需要继续拆分。
$$
\color{White}{
\blacktriangledown
}
$$
$$
\int \Bigg[\frac{A(x-1) + A + B}{(x – 1)^{2}} + \frac{Cx + D}{x^{2} + x + 1} \Bigg] \mathrm{d} x \Rightarrow
$$
$$
\int \Bigg[\frac{A}{(x – 1)} + \frac{A + B}{(x – 1)^{2}} + \frac{Cx + D}{x^{2} + x + 1} \Bigg] \mathrm{d} x \Rightarrow
$$
$$
令 E = A + B \Rightarrow
$$
$$
\int \Bigg[\frac{A}{(x – 1)} + \frac{E}{(x – 1)^{2}} + \frac{Cx + D}{x^{2} + x + 1} \Bigg] \mathrm{d} x \Rightarrow
$$
$$
\color{White}{
\blacktriangledown
}
$$
$$
\frac{A}{(x – 1)} + \frac{E}{(x – 1)^{2}} + \frac{Cx + D}{x^{2} + x + 1} = \frac{3x + 6}{(x – 1)^{2} (x^{2} + x + 1)} \Rightarrow
$$
$$
\color{White}{
\blacktriangledown
}
$$
$$
\frac{A(x – 1)^{2}(x^{2} + x + 1) + E(x-1)(x^{2} + x + 1) + (Cx + D)(x-1)^{3}}{(x – 1)^{3}(x^{2} + x + 1)} =
$$
$$
\frac{3x + 6}{(x – 1)^{2} (x^{2} + x + 1)} \Rightarrow
$$
$$
\color{White}{
\blacktriangledown
}
$$
$$
\frac{A(x – 1)(x^{2} + x + 1) + E(x^{2} + x + 1) + (Cx + D)(x-1)^{2}}{(x – 1)^{2}(x^{2} + x + 1)} =
$$
$$
\frac{3x + 6}{(x – 1)^{2} (x^{2} + x + 1)} \Rightarrow
$$
$$
\color{White}{
\blacktriangledown
}
$$
$$
A(x – 1)(x^{2} + x + 1) + E(x^{2} + x + 1) + (Cx + D)(x-1)^{2} =$$
$$
3x + 6 \Rightarrow
$$
$$
\color{White}{
\blacktriangledown
}
$$
$$
(Ax – A + E)(x^{2} + x + 1) + (Cx + D)(x^{2} + 1 – 2x) = 3x + 6 \Rightarrow
$$
$$
\color{White}{
\blacktriangledown
}
$$
$$
(Ax^{3} + Ax^{2} + Ax – Ax^{2} – Ax – A + Ex^{2} + Ex + E) +
$$
$$(Cx^{3} + Cx – 2Cx^{2} + Dx^{2} + D – 2Dx) =
$$
$$
3x + 6 \Rightarrow
$$
$$
\color{White}{
\blacktriangledown
}
$$
$$
(Ax^{3} – A + Ex^{2} + Ex + E) +
$$
$$(Cx^{3} + Cx – 2Cx^{2} + Dx^{2} + D – 2Dx) =
$$
$$
3x + 6 \Rightarrow
$$
$$
\color{White}{
\blacktriangledown
}
$$
$$
(A + C)x^{3} + (D + E – 2C)x^{2} + (C + E – 2D)x + (D + E – A) =
$$
$$
3x + 6 \Rightarrow
$$
$$
\color{White}{
\blacktriangledown
}
$$
$$
\left\{\begin{matrix}
A + C = 0;\\
D + E – 2C = 0;\\
C + E – 2D = 3;\\
D + E – A = 6.
\end{matrix}\right.
\Rightarrow
$$
$$
\left\{\begin{matrix}
A = -2;\\
E = 3;\\
C = 2;\\
D = 1.
\end{matrix}\right.
$$
于是：
$$
\int \frac{3x + 6}{(x – 1)^{2} (x^{2} + x + 1)} \mathrm{d} x =
$$
$$
\int \Bigg[ \frac{-2}{(x – 1)} + \frac{3}{(x – 1)^{2}} + \frac{2x + 1}{x^{2} + x + 1} \Bigg] \mathrm{d} x =
$$
$$
-2 \ln |x – 1| – \frac{3}{x-1} + \ln (x^{2} + x + 1) + C.
$$
其中 $C$ 为任意常数.