Subsection 5.3.1 分部积分法
分部积分法是计算不定积分的另一种重要方法, 它是针对乘积函数的求导法则导出的不定积分法. 在微分法中,设 \(u(x)\) 与 \(v(x)\) 是 \(x\) 的可微函数,有
\begin{equation*}
(u v)^{\prime}=u^{\prime} v+u v^{\prime} \text { 或 } u v^{\prime}=(u v)^{\prime}-u^{\prime} v .
\end{equation*}
对上述两式求不定积分得
\begin{equation*}
\displaystyle \int u v^{\prime} \mathrm{d} x=\displaystyle \int(u v)^{\prime} \mathrm{d} x-\displaystyle \int u^{\prime} v \mathrm{~d} x .
\end{equation*}
即
\begin{equation}
\displaystyle \int u v^{\prime} \mathrm{d} x=u v-\displaystyle \int u^{\prime} v \mathrm{~d} x,\tag{5.3.1}
\end{equation}
或
\begin{equation}
\displaystyle \int u \mathrm{~d} v=u v-\displaystyle \int v \mathrm{~d} u .\tag{5.3.2}
\end{equation}
以上两式称为分部积分公式. 分部积分公式表明, 求积分 \(\displaystyle \int u v^{\prime} \mathrm{d} x\) 有困难时, 可将积分分成 \(\displaystyle \int(u v)^{\prime} \mathrm{d} x\) 及 \(\displaystyle \int u^{\prime} v \mathrm{~d} x\) 两部分来积分, 且这两个积分都比较容易求解, 所以这种方法称为分部积分法. 下面通过例子说明如何运用分部积分公式.
Example 5.3.1.
例 1 求不定积分 \(\displaystyle \int x \sin x \mathrm{~d} x\text{.}\)
Solution.
解 被积表达式 \(x \sin x \mathrm{~d} x\) 可写成 \(x \mathrm{~d}(-\cos x)\) 的形式. 这里 \(u=x, v=-\cos x\text{.}\)由分部积分公式得
\begin{equation*}
\begin{aligned}
\displaystyle \int x \sin x \mathrm{~d} x & =\displaystyle \int x \mathrm{~d}(-\cos x)=x(-\cos x)-\displaystyle \int(-\cos x) \mathrm{d} x \\
& =-x \cos x+\displaystyle \int \cos x \mathrm{~d} x=-x \cos x+\sin x+C
\end{aligned}
\end{equation*}
注 如本例选 \(u=\sin x, v=\frac{x^{2}}{2}\text{,}\) 则 \(\displaystyle \int x \sin x \mathrm{~d} x=\displaystyle \int \sin x \mathrm{~d}\left(\frac{1}{2} x^{2}\right)=\frac{x^{2}}{2} \sin x-\) \(\displaystyle \int \frac{x^{2}}{2} \cos x \mathrm{~d} x\text{,}\) 而积分 \(\displaystyle \int \frac{x^{2}}{2} \cos x \mathrm{~d} x\) 比原来的积分还要复杂. 由此可见, 在运用分部积分法时,正确选择 \(u\) 和 \(v\) 很关键. 选取 \(u\) 和 \(\mathrm{d} v\) 一般考虑下面两点: (1) \(v\) 容易求出; (2) \(\displaystyle \int v \mathrm{~d} u\) 比 \(\displaystyle \int u \mathrm{~d} v\) 容易积分.
Example 5.3.2.
例 2 求不定积分 \(\displaystyle \int x \mathrm{e}^{x} \mathrm{~d} x\text{.}\)
Solution.
解 被积表达式 \(x \mathrm{e}^{x} \mathrm{~d} x\) 写成 \(x \mathrm{de}^{x}\) 的形式, 此时 \(u=x, v=\mathrm{e}^{x}\text{.}\) 由分部积分公式得
\begin{equation*}
\displaystyle \int x \mathrm{e}^{x} \mathrm{~d} x=\displaystyle \int x \mathrm{de}^{x}=x \mathrm{e}^{x}-\displaystyle \int \mathrm{e}^{x} \mathrm{~d} x=x \mathrm{e}^{x}-\mathrm{e}^{x}+C .
\end{equation*}
Example 5.3.3.
例 3 求不定积分 \(\displaystyle \int x^{2} \ln x \mathrm{~d} x\text{.}\)
Solution.
解 被积表达式 \(x^{2} \ln x \mathrm{~d} x\) 必须写成 \(\ln x \mathrm{~d}\left(\frac{1}{3} x^{3}\right)\) 的形式. 因为 \(\displaystyle \int \ln x \mathrm{~d} x\) 在基本积分表中找不到, 此时 \(u=\ln x, v=\frac{1}{3} x^{3}\text{,}\) 由分部积分公式得
\begin{equation*}
\begin{aligned}
\displaystyle \int x^{2} \ln x \mathrm{~d} x & =\displaystyle \int \ln x \mathrm{~d}\left(\frac{1}{3} x^{3}\right)=\left(\frac{1}{3} x^{3}\right) \ln x-\displaystyle \int \frac{1}{3} x^{3} \cdot \mathrm{d} \ln x \\
& =\frac{1}{3} x^{3} \ln x-\frac{1}{3} \displaystyle \int x^{3} \cdot \frac{1}{x} \mathrm{~d} x=\frac{1}{3} x^{3} \ln x-\frac{1}{9} x^{3}+C .
\end{aligned}
\end{equation*}
Example 5.3.4.
例 4 求不定积分 \(\displaystyle \int x \arctan x \mathrm{~d} x\text{.}\)
Solution.
解 被积表达式 \(x \arctan x \mathrm{~d} x\) 可写成 \(\arctan x \mathrm{~d}\left(\frac{1}{2} x^{2}\right)\) 的形式. 此时 \(u=\arctan x\text{,}\) \(v=\frac{1}{2} x^{2}\text{,}\) 由分部积分公式得
\begin{equation*}
\begin{aligned}
\displaystyle \int x \arctan x \mathrm{~d} x & =\displaystyle \int \arctan x \mathrm{~d}\left(\frac{1}{2} x^{2}\right) \\
& =\frac{1}{2} x^{2} \arctan x-\displaystyle \int \frac{1}{2} x^{2} \mathrm{~d}(\arctan x) \\
& =\frac{1}{2} x^{2} \arctan x-\frac{1}{2} \displaystyle \int \frac{x^{2}}{1+x^{2}} \mathrm{~d} x \\
& =\frac{1}{2} x^{2} \arctan x-\frac{1}{2} \displaystyle \int \frac{x^{2}+1-1}{1+x^{2}} \mathrm{~d} x \\
& =\frac{1}{2} x^{2} \arctan x-\frac{1}{2}\left(\displaystyle \int \mathrm{d} x-\displaystyle \int \frac{1}{1+x^{2}} \mathrm{~d} x\right) \\
& =\frac{1}{2} x^{2} \arctan x-\frac{1}{2} x+\frac{1}{2} \arctan x+C .
\end{aligned}
\end{equation*}
Example 5.3.5.
例 5 求不定积分 \(\displaystyle \int \mathrm{e}^{a x} \cos b x \mathrm{~d} x\) 与 \(\displaystyle \int \mathrm{e}^{a x} \sin b x \mathrm{~d} x \quad(a \neq 0)\text{.}\)
Solution.
解 \(\quad \displaystyle \int \mathrm{e}^{a x} \cos b x \mathrm{~d} x=\frac{1}{a} \mathrm{e}^{a x} \cos b x+\frac{b}{a} \displaystyle \int \mathrm{e}^{a x} \sin b x \mathrm{~d} x\text{;}\)
\begin{equation*}
\displaystyle \int \mathrm{e}^{a x} \sin b x \mathrm{~d} x=\frac{1}{a} \mathrm{e}^{a x} \sin b x-\frac{b}{a} \displaystyle \int \mathrm{e}^{a x} \cos b x \mathrm{~d} x .
\end{equation*}
可见这两个积分可以相互表示出来, 若用加减消去法, 则有
\begin{equation*}
\displaystyle \int \mathrm{e}^{a x} \cos b x \mathrm{~d} x=\frac{b \sin b x+a \cos b x}{a^{2}+b^{2}} \mathrm{e}^{a x}+C
\end{equation*}
与
\begin{equation*}
\displaystyle \int \mathrm{e}^{a x} \sin b x \mathrm{~d} x=\frac{a \sin b x-b \cos b x}{a^{2}+b^{2}} \mathrm{e}^{a x}+C .
\end{equation*}
Example 5.3.6.
例 6 求不定积分 \(\displaystyle \int \sin (\ln x) \mathrm{d} x\text{.}\)
Solution.
解 这时设 \(u(x)=\sin (\ln x), \mathrm{d} x=\mathrm{d} v\text{,}\) 即
\begin{equation*}
\begin{aligned}
\displaystyle \int \sin (\ln x) \mathrm{d} x & =x \sin (\ln x)-\displaystyle \int x \mathrm{~d}[\sin (\ln x)] \\
& =x \sin (\ln x)-\displaystyle \int x \cos (\ln x) \frac{1}{x} \mathrm{~d} x \\
& =x[\sin (\ln x)-\cos (\ln x)]-\displaystyle \int \sin (\ln x) \mathrm{d} x .
\end{aligned}
\end{equation*}
所以
\begin{equation*}
\displaystyle \int \sin (\ln x) \mathrm{d} x=\frac{1}{2} x[\sin (\ln x)-\cos (\ln x)]+C .
\end{equation*}
Example 5.3.8.
例 7 求不定积分 \(J_{n}=\displaystyle \int \frac{1}{\left(a^{2}+x^{2}\right)^{n}} \mathrm{~d} x \quad(a \neq 0)\text{.}\)
Solution.
解 \(J_{n}=\displaystyle \int \frac{1}{\left(a^{2}+x^{2}\right)^{n}} \mathrm{~d} x=\frac{x}{\left(x^{2}+a^{2}\right)^{n}}+2 n \displaystyle \int \frac{x^{2}}{\left(x^{2}+a^{2}\right)^{n+1}} \mathrm{~d} x\)
\begin{equation*}
\begin{aligned}
& =\frac{x}{\left(x^{2}+a^{2}\right)^{n}}+2 n \displaystyle \int \frac{\left(x^{2}+a^{2}\right)-a^{2}}{\left(x^{2}+a^{2}\right)^{n+1}} \mathrm{~d} x \\
& =\frac{x}{\left(x^{2}+a^{2}\right)^{n}}+2 n \displaystyle \int \frac{1}{\left(x^{2}+a^{2}\right)^{n}} \mathrm{~d} x-2 n a^{2} \displaystyle \int \frac{1}{\left(x^{2}+a^{2}\right)^{n+1}} \mathrm{~d} x \\
& =\frac{x}{\left(x^{2}+a^{2}\right)^{n}}+2 n J_{n}-2 n a^{2} J_{n+1},
\end{aligned}
\end{equation*}
则
\begin{equation*}
J_{n+1}=\frac{1}{2 n a^{2}} \frac{x}{\left(x^{2}+a^{2}\right)^{n}}+\frac{2 n-1}{2 n} \cdot \frac{1}{a^{2}} J_{n} \quad(n \geqslant 1) .
\end{equation*}
这个公式称为 \(J_{n}\) 的递推公式.
当 \(n=1\) 时,有 \(J_{1}=\displaystyle \int \frac{1}{x^{2}+a^{2}} \mathrm{~d} x=\frac{1}{a} \arctan \frac{x}{a}+C\text{.}\)
Example 5.3.9.
例 8 求不定积分 \(I_{n}=\displaystyle \int(\ln x)^{n} \mathrm{~d} x\text{.}\)
Solution.
解 \(I_{n}=\displaystyle \int(\ln x)^{n} \mathrm{~d} x=x(\ln x)^{n}-n \displaystyle \int(\ln x)^{n-1} \mathrm{~d} x=x(\ln x)^{n}-n I_{n-1}\) ,
有
\begin{equation*}
I_{n}=x(\ln x)^{n}-n I_{n-1} \quad(n \geqslant 2) .
\end{equation*}
当 \(n=1\) 时,有
\begin{equation*}
I_{1}=\displaystyle \int \ln x \mathrm{~d} x=x \ln x-x+C .
\end{equation*}
在计算积分时,有时要把不定积分的分部积分法与不定积分的换元法结合使用.
Example 5.3.10.
例 9 求不定积分 \(\displaystyle \int \sec ^{3} x \mathrm{~d} x\text{.}\)
Solution.
解 \(\displaystyle \int \sec ^{3} x \mathrm{~d} x=\displaystyle \int \sec x \sec ^{2} x \mathrm{~d} x=\displaystyle \int \sec x \mathrm{~d}(\tan x)=\sec x \tan x-\displaystyle \int \tan x \mathrm{~d}(\sec x)\)
\begin{equation*}
\begin{aligned}
& =\sec x \tan x-\displaystyle \int \tan ^{2} x \sec x \mathrm{~d} x \\
& =\sec x \tan x-\displaystyle \int\left(\sec ^{2} x-1\right) \sec x \mathrm{~d} x \\
& =\sec x \tan x-\displaystyle \int \sec ^{3} x \mathrm{~d} x+\displaystyle \int \sec x \mathrm{~d} x,
\end{aligned}
\end{equation*}
所以 \(\displaystyle \int \sec ^{3} x \mathrm{~d} x=\frac{1}{2}\left[\sec x \tan x+\displaystyle \int \sec x \mathrm{~d} x\right]\)
\begin{equation*}
=\frac{1}{2}[\sec x \tan x+\ln |\sec x+\tan x|]+C \text {. }
\end{equation*}
Example 5.3.11.
例 10 求不定积分 \(\displaystyle \int \mathrm{e}^{\sqrt{x}} \mathrm{~d} x\text{.}\)
Solution.
解 令 \(\sqrt{x}=t\text{,}\) 则 \(x=t^{2}, \mathrm{~d} x=2 t \mathrm{~d} t\text{,}\)
于是 \(\quad \displaystyle \int \mathrm{e}^{\sqrt{x}} \mathrm{~d} x=2 \displaystyle \int t \mathrm{e}^{t} \mathrm{~d} t=2 \mathrm{e}^{t}(t-1)+C=2 \mathrm{e}^{\sqrt{x}}(\sqrt{x}-1)+C\text{.}\)
Example 5.3.12.
例 11 求不定积分 \(\displaystyle \int \cos \sqrt{1+x} \mathrm{~d} x\text{.}\)
Solution.
解 令 \(\sqrt{1+x}=t\text{,}\) 则 \(x=t^{2}-1, \mathrm{~d} x=2 t \mathrm{~d} t\text{,}\)
于是 \(\quad \displaystyle \int \cos \sqrt{1+x} \mathrm{~d} x=\displaystyle \int \cos t \cdot 2 t \mathrm{~d} t=2 \displaystyle \int t \mathrm{~d}(\sin t)=2 t \sin t-2 \displaystyle \int \sin t \mathrm{~d} t\)
\begin{equation*}
\begin{aligned}
& =2 t \sin t+2 \cos t+C \\
& =2 \sqrt{1+x} \sin \sqrt{1+x}+2 \cos \sqrt{1+x}+C .
\end{aligned}
\end{equation*}
Example 5.3.13.
例 12 求不定积分 \(\displaystyle \int \frac{\arcsin \mathrm{e}^{x}}{\mathrm{e}^{x}} \mathrm{~d} x\text{.}\)
Solution.
解 令 \(\mathrm{e}^{x}=t>0\text{,}\) 则 \(x=\ln t, \mathrm{~d} x=\frac{1}{t} \mathrm{~d} t\text{,}\) 有
\begin{equation*}
\begin{aligned}
\displaystyle \int \frac{\arcsin \mathrm{e}^{x}}{\mathrm{e}^{x}} \mathrm{~d} x & =\displaystyle \int \arcsin t \mathrm{~d}\left(-\frac{1}{t}\right)=-\frac{1}{t} \arcsin t+\displaystyle \int \frac{1}{t \sqrt{1-t^{2}}} \mathrm{~d} t \\
& =-\frac{1}{t} \arcsin t+\displaystyle \int \frac{1}{t^{2} \sqrt{\left(\frac{1}{t}\right)^{2}-1}} \mathrm{~d} t \\
& =-\frac{1}{t} \arcsin t-\displaystyle \int \frac{1}{\sqrt{\left(\frac{1}{t}\right)^{2}-1}} \mathrm{~d}\left(\frac{1}{t}\right) \\
& =-\frac{1}{t} \arcsin t-\ln \left[\frac{1}{t}+\sqrt{\left(\frac{1}{t}\right)^{2}-1}\right]+C \\
& =-\frac{1}{t} \arcsin t+\ln t-\ln \left(1+\sqrt{1-t^{2}}\right)+C \\
& =-\mathrm{e}^{-x} \arcsin \mathrm{e}^{x}+x-\ln \left(1+\sqrt{1-\mathrm{e}^{2 x}}\right)+C .
\end{aligned}
\end{equation*}
Example 5.3.14.
例 13 求不定积分 \(\displaystyle \int \frac{\arcsin x}{x^{2}} \mathrm{~d} x\text{.}\)
Solution.
解 \(\displaystyle \int \frac{\arcsin x}{x^{2}} \mathrm{~d} x=-\displaystyle \int \arcsin x \mathrm{~d}\left(\frac{1}{x}\right)=-\frac{1}{x} \arcsin x+\displaystyle \int \frac{1}{x} \mathrm{~d}(\arcsin x)\)
\begin{equation*}
=-\frac{1}{x} \arcsin x+\displaystyle \int \frac{\mathrm{d} x}{x \sqrt{1-x^{2}}} .
\end{equation*}
对上式右端第二个积分, 令 \(x=\sin t\text{,}\)于是
\begin{equation*}
\begin{aligned}
\displaystyle \int \frac{\mathrm{d} x}{x \sqrt{1-x^{2}}} & =\displaystyle \int \frac{\cos t \mathrm{~d} t}{\sin t \cos t}=\displaystyle \int \frac{1}{\sin t} \mathrm{~d} t=\ln |\csc t-\cot t|+C \\
& =\ln \left|\frac{1}{x}-\frac{\sqrt{1-x^{2}}}{x}\right|+C,
\end{aligned}
\end{equation*}
所以
\begin{equation*}
\displaystyle \int \frac{\arcsin x}{x^{2}} \mathrm{~d} x=-\frac{1}{x} \arcsin x+\ln \left|\frac{1-\sqrt{1-x^{2}}}{x}\right|+C .
\end{equation*}
