2025年考研数二第17题解析：定积分的计算、因式分解

一、题目

二、解析

解法 1：先做代换，再做因式分解

$$
\begin{aligned}
& \ \int_{0}^{1} \frac{1}{\left( x+1 \right)\left( x^{2}-2x+2 \right)} \mathrm{~d}x \\ \\
= & \ \int_{0}^{1} \frac{1}{\left( x+1 \right)\left[ \left( x-1 \right)^{2}+1 \right]} \mathrm{~d}x \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{t=x-1} \\ \\
= & \ \int_{-1}^{0} \frac{1}{\left( t+2 \right)\left( t^{2}+1 \right)} \mathrm{~d}t \\ \\
= & \ \int_{-1}^{0} \left( \frac{A}{t+2} + \frac{Bt + C}{t^{2} + 1} \right) \mathrm{~d} t \\ \\
= & \ \int_{-1}^{0} \frac{\left( A+B \right)t^{2} + \left( 2B + C \right) t + A + 2C}{\left( t+2 \right)\left( t^{2}+1 \right)} \mathrm{~d} t \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\begin{cases}
A + B = 0 \\
2B + C = 0 \\
A + 2C = 1
\end{cases} \leadsto \begin{cases}
A = 1 \\
B = -1 \\
C = 2
\end{cases}} \\ \\
= & \ \frac{1}{5} \int_{-1}^{0}\left( \frac{1}{t+2}+\frac{-t+2}{t^{2}+1} \right) \mathrm{~d}t \\ \\
= & \ \left.\frac{1}{5}\left[ \ln \left( t+2 \right)-\frac{1}{2}\ln \left( t^{2}+1 \right)+2\arctan x \right] \right|_{-1}^{0} \\ \\
= & \ \frac{1}{5} \left[ \ln 2-\left( -\frac{1}{2}\ln 2-\frac{\pi}{2} \right) \right] \\ \\
= & \ \frac{1}{5}\left( \frac{3}{2}\ln 2+\frac{\pi}{2} \right)
\end{aligned}
$$

解法 2：只做因式分解

$$
\begin{aligned}
& \ \int_{0}^{1}\frac{1}{\left(x+1\right)\left(x^{2}-2x+2\right)} \mathrm{~d}x \\ \\
= & \ \int_{0}^{1}\left(\frac{A}{x+1}+\frac{Bx+C}{x^{2}-2x+2}\right) \mathrm{~d}x \\ \\
= & \ \int_{0}^{1} \left( \frac{\left( A+B \right)x^{2} + \left( -2 A + B + C \right) x + 2A + C}{\left( 1+x \right) \left( x^{2} – 2x + 2 \right)} \right) \mathrm{~d} x \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\begin{cases}
A+B=0 \\
-2A+B+C=0 \\
2A+C=1
\end{cases} \leadsto \begin{cases}
A = \frac{1}{5} \\
B = \frac{-1}{5} \\
C = \frac{3}{5}
\end{cases} } \\ \\
= & \ \int_{0}^{1}\left(\frac{\frac{1}{5}}{x+1}+\frac{-\frac{1}{5}x+\frac{3}{5}}{x^{2}-2x+2}\right) \mathrm{~d}x \\ \\
= & \ \left.\frac{1}{5}\ln\left|1+x\right|\right|_{0}^{1} – \left.\frac{1}{10}\ln\left|x^{2}-2x+2\right|\right|_{0}^{1} + \left.\frac{2}{5}\arctan\left(x-1\right)\right|_{0}^{1} \\ \\
= & \ \frac{3}{10}\ln 2 + \frac{1}{10}\pi
\end{aligned}
$$

---