考研数学常用积分之：含有 $ax$ $+$ $b$ 的积分

一、前言

在本文中，「荒原之梦考研数学」将为同学们总结整理被积函数中含有 “$ax$ $+$ $b$” 以及相关变形形式的积分，这些不是基础的积分公式，也不是一般的习题，但可以作为同学们对积分解题方法的积累。

二、正文

§2.1 没有根号

$$\begin{array}{rl}\mathbf{\star }& \int \frac{1}{ax+b}\mathrm{d}x\\ \\ =& \frac{1}{a}\ln |ax+b|+C\\ \\ \\ \mathbf{\star }& \int \frac{x}{ax+b}\mathrm{d}x\\ \\ =& \frac{1}{a}(ax+b-b\ln |cx+b|)+C\\ \\ \\ \mathbf{\star }& \int \frac{x}{(ax+b{)}^2}\mathrm{d}x\\ \\ =& \frac{1}{a^2}(\ln |ax+b|+\frac{b}{ax+b})+C\\ \\ \\ \mathbf{\star }& \int \frac{x^2}{ax+b}\mathrm{d}x\\ \\ =& \frac{1}{a^3}[\frac{1}{2}(ax+b{)}^2-2b(ax+b)+b^2\ln |ax+b|]+C\\ \\ \\ \mathbf{\star }& \int \frac{x^2}{(ax+b{)}^2}\mathrm{d}x\\ \\ =& \frac{1}{a^3}(ax+b-2b\ln |ax+b|-\frac{b^2}{ax+b})+C\\ \\ \\ \mathbf{\star }& \int \frac{1}{x(ax+b)}\mathrm{d}x\\ \\ =& \frac{–1}{b}\ln \left|\frac{ax+b}{x}\right|+C\\ \\ \\ \mathbf{\star }& \int \frac{1}{x^2(ax+b)}\mathrm{d}x\\ \\ =& \frac{-1}{bx}+\frac{a}{b^2}\ln \left|\frac{ax+b}{x}\right|+C\\ \\ \\ \mathbf{\star }& \int \frac{1}{x(ax+b{)}^2}\mathrm{d}x\\ \\ =& \frac{1}{b(ax+b)}-\frac{1}{b^2}\ln \left|\frac{ax+b}{x}\right|+C\\ \\ \\ \mathbf{\star }& \int (ax+b{)}^k\mathrm{d}x\\ \\ =& \frac{1}{a(k+1)}(ax+b{)}^{k+1}+C(k\ne -1)\end{array}$$

§2.2 有根号

$$\begin{array}{rl}\mathbf{\star }& \int \sqrt{ax+b}\mathrm{d}x\\ \\ =& \frac{2}{3a}\sqrt{(ax+b{)}^3}+C\\ \\ \\ \mathbf{\star }& \int x\sqrt{ax+b}\mathrm{d}x\\ \\ =& \frac{2}{15a^2}(3ax-2b)\sqrt{(ax+b{)}^3}+C\\ \\ \\ \mathit{◊}& \int x^2\sqrt{ax+b}\mathrm{d}x\\ \\ =& \frac{2}{105a^3}(15a^2{x}^2-12abx+8b^2)\sqrt{(ax+b{)}^3}+C\\ \\ \\ \mathbf{\star }& \int \frac{x}{\sqrt{ax+b}}\mathrm{d}x\\ \\ =& \frac{2}{3a^2}(ax–2b)\sqrt{ax+b}+C\\ \\ \\ \mathit{◊}& \int \frac{x^2}{\sqrt{ax+b}}\mathrm{d}x\\ \\ =& \frac{2}{15a^3}(3a^2{x}^2–4abx+8b^2)\sqrt{ax+b}+C\\ \\ \\ \mathbf{\star }& \int \frac{1}{x\sqrt{ax+b}}\mathrm{d}x\\ \\ =& \{\begin{array}{ll}\frac{1}{\sqrt{b}}\ln \left|\frac{\sqrt{ax+b}-\sqrt{b}}{\sqrt{ax+b}+\sqrt{b}}\right|+C,& b>0\\ \\ \frac{2}{\sqrt{-b}}\arctan \sqrt{\frac{ax+b}{-b}}+C,& b<0\end{array}\\ \\ \\ \mathit{◊}& \int \frac{1}{x^2\sqrt{ax+b}}\mathrm{d}x\\ \\ =& -\sqrt{\frac{ax+b}{bx}}-\frac{a}{2b}\int \frac{1}{x\sqrt{ax+b}}\mathrm{d}x\\ \\ =& -\sqrt{\frac{ax+b}{bx}}-\frac{a}{2b}\{\begin{array}{ll}\frac{1}{\sqrt{b}}\ln \left|\frac{\sqrt{ax+b}-\sqrt{b}}{\sqrt{ax+b}+\sqrt{b}}\right|+C,& b>0\\ \\ \frac{2}{\sqrt{-b}}\arctan \sqrt{\frac{ax+b}{-b}}+C,& b<0\end{array}\\ \\ \\ \mathit{◊}& \int \sqrt{\frac{ax+b}{x}}\mathrm{d}x\\ \\ =& 2\sqrt{ax+b}+b\int \frac{1}{x\sqrt{ax+b}}\mathrm{d}x\\ \\ =& 2\sqrt{ax+b}+b\{\begin{array}{ll}\frac{1}{\sqrt{b}}\ln \left|\frac{\sqrt{ax+b}-\sqrt{b}}{\sqrt{ax+b}+\sqrt{b}}\right|+C,& b>0\\ \\ \frac{2}{\sqrt{-b}}\arctan \sqrt{\frac{ax+b}{-b}}+C,& b<0\end{array}\\ \\ \\ \mathit{◊}& \int \frac{\sqrt{ax+b}}{x^2}\mathrm{d}x\\ \\ =& \frac{-\sqrt{ax+b}}{x}+\frac{a}{2}\int \frac{1}{x\sqrt{ax+b}}\mathrm{d}x\\ \\ =& \frac{-\sqrt{ax+b}}{x}+\frac{a}{2}\{\begin{array}{ll}\frac{1}{\sqrt{b}}\ln \left|\frac{\sqrt{ax+b}-\sqrt{b}}{\sqrt{ax+b}+\sqrt{b}}\right|+C,& b>0\\ \\ \frac{2}{\sqrt{-b}}\arctan \sqrt{\frac{ax+b}{-b}}+C,& b<0\end{array}\end{array}$$

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