不定积分题集
紫钦
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个人记录
前言
本篇 blog 主要记载一些我遇到的关于不定积分的比较好或者常见的题目。
持续更新中。
题目
- 求不定积分 \int\dfrac{x^2}{1+x^4}dx 。
解：此为有理函数的不定积分，可使用有理函数不定积分的通用方式，即化为 7 个基本有理函数的组合。
法一：有理函数法，
\ \ \ \ 1+x^4
=(1+2x^2+x^4)-2x^2
=(1+x^2)^2-(\sqrt2x)^2
于是，
$\ \ \ \ \int\dfrac{x^2}{1+x^4}dx
=\int\dfrac{x^2}{(1+\sqrt2x+x^2)(1-\sqrt2x+x^2)}dx
=\int(\dfrac{-\frac{\sqrt2}{4}x}{1+\sqrt2x+x^2}+\dfrac{\frac{\sqrt2}{4}x}{1-\sqrt2x+x^2})dx
=-\dfrac{\sqrt2}{8}\int\dfrac{2x}{1+\sqrt2x+x^2}dx+\dfrac{\sqrt2}{8}\int\dfrac{2x}{1-\sqrt2x+x^2}dx
=-\dfrac{\sqrt2}{8}\int(\dfrac{2x+\sqrt2}{1+\sqrt2x+x^2}-\dfrac{\sqrt2}{1+\sqrt2x+x^2})dx+\dfrac{\sqrt2}{8}\int(\dfrac{2x-\sqrt2}{1-\sqrt2x+x^2}+\dfrac{\sqrt2}{1-\sqrt2x+x^2})dx
=\dfrac{\sqrt2}{8}\ln\dfrac{1-\sqrt2x+x^2}{1+\sqrt2x+x^2}+\dfrac{\sqrt2}{8}\int\dfrac{\sqrt2}{1+\sqrt2x+x^2}dx+\dfrac{\sqrt2}{8}\int\dfrac{\sqrt2}{1-\sqrt2x+x^2}dx
=\dfrac{\sqrt2}{8}\ln\dfrac{1-\sqrt2x+x^2}{1+\sqrt2x+x^2}+\dfrac{1}{4}\int\dfrac{1}{(x+\frac{\sqrt2}{2})^2+\frac12}dx+\dfrac{1}{4}\int\dfrac{1}{(x-\frac{\sqrt2}{2})^2+\frac12}dx
法二：凑微分法，
$\ \ \ \ \int\dfrac{x^2}{1+x^4}dx
=\dfrac12\int\dfrac{(x^2-1)+(x^2+1)}{1+x^4}dx
=\dfrac12\int\dfrac{x^2-1}{1+x^4}dx+\dfrac12\int\dfrac{x^2+1}{1+x^4}dx
=\dfrac12\int\dfrac{1-\frac{1}{x^2}}{\frac{1}{x^2}+x^2}dx+\dfrac12\int\dfrac{1+\frac{1}{x^2}}{\frac{1}{x^2}+x^2}dx
=\dfrac12\int\dfrac{1}{\frac{1}{x^2}+x^2}d(x+\dfrac{1}{x})+\dfrac12\int\dfrac{1}{\frac{1}{x^2}+x^2}d(x-\dfrac{1}{x})
=\dfrac12\int\dfrac{1}{(x+\frac{1}{x})^2-2}d(x+\dfrac{1}{x})+\dfrac12\int\dfrac{1}{(x-\frac{1}{x})^2+2}d(x-\dfrac{1}{x})
=\dfrac{\sqrt2}{8}\ln\dfrac{x+\frac{1}{x}-\sqrt2}{x+\frac{1}{x}+\sqrt2}+\dfrac{\sqrt2}{4}\arctan\dfrac{x-\frac{1}{x}}{\sqrt2}+C
=\dfrac{\sqrt2}{8}\ln\dfrac{x^2-\sqrt2x+1}{x^2+\sqrt2x+1}+\dfrac{\sqrt2}{4}\arctan\dfrac{x^2-1}{\sqrt2x}+C
相较而言，法二更容易。
- 求不定积分 \int\arctan xdx。
解：分部积分法。
\ \ \ \ \int\arctan xdx
=x\arctan x-\int\dfrac{x}{1+x^2}dx
=x\arctan x-\dfrac12\int\dfrac{1}{1+x^2}d(1+x^2)
=x\arctan x-\dfrac12\ln(1+x^2)
记住这个结论。