一、题目
$\int_{0}^{1} \frac{2 x + 3}{x^{2} – x + 1} \mathrm{~d} x = \underline{\qquad \qquad}$.
二、解析
$$
\begin{aligned}
& \ \int_{0}^{1} \frac{2 x + 3}{x^{2} – x + 1} \mathrm{~d} x \\ \\
= & \ \int_{0}^{1} \frac{2 x – 1 + 4}{x^{2} – x + 1} \mathrm{~d} x \\ \\
= & \ \int_{0}^{1} \frac{1}{x^{2} – x + 1} \mathrm{~d} \left(x^{2} – x + 1\right) + \int_{0}^{1} \frac{4}{x^{2} – x + 1} \mathrm{~d} x \\ \\
= & \ \left.\ln \left(x^{2} – x + 1\right)\right|_{0}^{1} + 4 \int_{0}^{1} \frac{1}{x^{2} – x + 1} \mathrm{~d} x \\ \\
= & \ 0 + 4 \int_{0}^{1} \frac{1}{x^{2} – x + 1} \mathrm{~d} x \\ \\
= & \ 4 \int_{0}^{1} \frac{1}{\left(x – \frac{1}{2}\right)^{2} + \frac{3}{4}} \mathrm{~d} x \\ \\
= & \ 4 \int_{0}^{1} \frac{1}{\left(x – \frac{1}{2}\right)^{2} + \frac{3}{4}} \mathrm{~d} \left(x – \frac{1}{2}\right) \\ \\
= & \ 4 \int_{0}^{1} \frac{1}{\left(x – \frac{1}{2}\right)^{2} + \left(\frac{\sqrt{3}}{2}\right)^{2}} \mathrm{~d} \left(x – \frac{1}{2}\right) \\ \\
= & \ 4 \cdot \frac{4}{3} \cdot \frac{\sqrt{3}}{2} \cdot \int_{0}^{1} \frac{1}{\left( \frac{x – \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)^{2} + 1} \mathrm{~d} \left( \frac{x – \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) \\ \\
= & \ \left. 4 \cdot \frac{2}{\sqrt{3}} \arctan \frac{x – \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right|_{0}^{1} \\ \\
= & \ 4 \cdot \frac{2}{\sqrt{3}} \left( \arctan \frac{\sqrt{3}}{3} – \arctan \left( – \frac{\sqrt{3}}{3} \right) \right) \\ \\
= & \ 4 \cdot \frac{2}{\sqrt{3}} \left( \frac{\pi}{6} – \left( \frac{- \pi}{6} \right) \right) \\ \\
= & \ 4 \cdot \frac{2}{\sqrt{3}} \cdot \frac{\pi}{3} \\ \\
= & \ \textcolor{lightgreen}{ \frac{8 \sqrt{3}}{9} \pi }
\end{aligned}
$$
三、题型总结
本题中的被积函数为有理函数，形如 $\frac{M x + N}{x^{2} + p x + q}$, 其中被积函数的分子为一次多项式，分母为没有实数根的二次三项式. 对此类积分，一个常见的解题思路，就是将其拆分为能够使用 $\int \frac{1}{x} \mathrm{~d} x = \ln x + C$ 和 $\int \frac{1}{1+x^{2}} \mathrm{~d} x = \arctan x + C$ 这两个积分公式的形式：
$$
\begin{aligned}
\int \frac{M x + N}{x^{2} + p x + q} \mathrm{~d} x & = \int \frac{\left(2 x + p\right) \cdot \frac{M}{2}}{x^{2} + p x + q} \mathrm{~d} x + \int \frac{N – \frac{p M}{2}}{\left(x + \frac{p}{2}\right)^{2} + q – \frac{p^{2}}{4}} \mathrm{~d} x \\ \\
& = \int \frac{\left(2 x + p\right) \cdot \frac{M}{2}}{x^{2} + p x + q} \mathrm{~d} x + \int \frac{N – \frac{p M}{2}}{\left(x + \frac{p}{2}\right)^{2} + q – \frac{p^{2}}{4}} \mathrm{~d} x \\ \\
& = \frac{M}{2} \int \frac{1}{x^{2} + p x + q} \mathrm{~d} \left( x^{2} + px + q \right) + \int \frac{N – \frac{p M}{2}}{\left(x + \frac{p}{2}\right)^{2} + q – \frac{p^{2}}{4}} \mathrm{~d} \left(x + \frac{p}{2}\right) \\ \\
& = \frac{M}{2} \int \frac{1}{x^{2} + p x + q} \mathrm{~d} \left( x^{2} + px + q \right) + \left( q-\frac{p^{2}}{4} \right)^{-1} \cdot \int \frac{N – \frac{p M}{2}}{\frac{\left(x + \frac{p}{2}\right)^{2}}{q-\frac{p^{2}}{4}} + 1} \mathrm{~d} \left(x + \frac{p}{2}\right) \\ \\
& = \frac{M}{2} \int \frac{1}{x^{2} + p x + q} \mathrm{~d} \left( x^{2} + px + q \right) + \left[ \frac{1}{4} \left( 4q-p^{2} \right) \right]^{-1} \cdot \int \frac{N – \frac{p M}{2}}{\frac{ \frac{1}{4} \left(2x + p \right)^{2}}{\frac{1}{4} \left( 4q-p^{2} \right)} + 1} \mathrm{~d} \left(x + \frac{p}{2}\right) \\ \\
& = \frac{M}{2} \int \frac{1}{x^{2} + p x + q} \mathrm{~d} \left( x^{2} + px + q \right) + \frac{4}{4q-p^{2}} \cdot \int \frac{N – \frac{p M}{2}}{\frac{ \left(2x + p \right)^{2}}{ \left( 4q-p^{2} \right)} + 1} \mathrm{~d} \left(x + \frac{p}{2}\right) \\ \\
& = \frac{M}{2} \int \frac{1}{x^{2} + p x + q} \mathrm{~d} \left( x^{2} + px + q \right) + \frac{4}{4q-p^{2}} \cdot \int \frac{N – \frac{p M}{2}}{\left( \frac{ 2x + p }{ \sqrt{4q-p^{2}}} \right)^{2} + 1} \mathrm{~d} \left(x + \frac{p}{2}\right) \\ \\
& = \frac{M}{2} \int \frac{1}{x^{2} + p x + q} \mathrm{~d} \left( x^{2} + px + q \right) + \frac{4}{4q-p^{2}} \cdot \frac{\sqrt{4q-p^{2}}}{2} \cdot \left( N – \frac{p M}{2} \right) \int \frac{1}{\left( \frac{ 2x + p }{ \sqrt{4q-p^{2}}} \right)^{2} + 1} \mathrm{~d} \left( \frac{ 2x + p }{ \sqrt{4q-p^{2}}} \right) \\ \\
& = \frac{M}{2} \int \frac{1}{x^{2} + p x + q} \mathrm{~d} \left( x^{2} + px + q \right) + \frac{4}{4q-p^{2}} \cdot \frac{\sqrt{4q-p^{2}}}{2} \cdot \left( N – \frac{p M}{2} \right) \int \frac{1}{\left( \frac{ 2x + p }{ \sqrt{4q-p^{2}}} \right)^{2} + 1} \mathrm{~d} \left( \frac{ 2x + p }{ \sqrt{4q-p^{2}}} \right) \\ \\
& = \frac{M}{2} \ln \left(x^{2} + p x + q\right) + \frac{2 \left(N – \frac{p M}{2}\right)}{\sqrt{4 q – p^{2}}}
\arctan \frac{2 x + p}{\sqrt{4 q – p^{2}}} + C
\end{aligned}
$$
四、相关知识点
[1]. 常用三角函数的取值对照表