学了积分当然要做一些很水的题目来长自信,今天就来整理一下
1. \[\int {\frac{{\sqrt {{x^2} + 1} – \sqrt {{x^2} – 1} }}{{\sqrt {{x^4} – 1} }}} {\text{d}}x\]
这题一看就是标准毒瘤题是吧…23333
事实上挺简单的啦,只要你没想着一个劲地换元,把分母拆开来算即可
答案:\[\ln |x + \sqrt {{x^2} + 1} | – \ln |x + \sqrt {{x^2} – 1} | + C\]

2. \[\int {\frac{{{{\sec }^2}(x)}}{{9 + {{\tan }^2}(x)}}} {\text{d}}x\]
这题一眼化切为弦,然后就没了
答案: \[\frac{{\text{1}}}{{\text{3}}}\arctan \left( {\frac{{\tan (x)}}{3}} \right) + C,x \ne \frac{\pi }{2} + k\pi ,k \in \mathbb{Z}\]
3. \[\int {\frac{{{\text{d}}x}}{{\sqrt {x – 1} + \sqrt {{{(x – 1)}^3}} }}{\text{d}}x} \]
这题简直太裸了吧,谁要是再看不出来换元积分他可以退学了
有一个注意的地方就是这里我们需要分母有理化,然后还要再换一次
作换元\( u=x-1\)
\[
\begin{aligned}
原式&=\int {\frac{1}{{\sqrt {{u^3}} + \sqrt u }}} {\mkern 1mu} {\text{d}}u \\
&= \int \frac{\sqrt{u^3}-\sqrt{u}}{u^3-u} \, \text{d}u \\
&= \int \frac{u^{3/2} \left(-u^3+u^2-u+1\right)}{u^2-u^6} \, \text{d}u
\end{aligned}
\]
这有什么好处呢?我们可以再往下看
继续作换元\( s = \sqrt{u}\),就有\({\text{d}}s = \frac{1}{{2\sqrt u }}{\text{d}}u\)
\[\begin{aligned}
原式&=2 \int \frac{1}{s^2+1} \, \text{d}s \\
&= 2\arctan (s) + C \\
&= 2\arctan \left( {\sqrt u } \right) + C \\
&= 2\arctan \left( {\sqrt {x – 1} } \right) + C
\end{aligned}
\]
4. \[ \int \frac{x}{\tan \left(x^2+1\right)} \, \text{d}x \]
这题蠢爆了
官方解答给的什么鬼玩意我都不想看
还是换元,换换换作换元\( u = \arctan(x) \),即得\( \displaystyle \text{d} u = \frac{1}{x^2+1}\text{d}x\)
原式就等于\( \displaystyle \int u \text{d} u \)
然后就得到\( \dfrac{u^2}{2}+C\)
然后没了
好了,最后一道题了,来难(烦)一些的吧…
5. \[ \int \frac{3 x^5-10 x^4+21 x^3-42 x^2+36 x-32}{x \left(x^2+1\right) \left(x^2+4\right)^2} \, \text{d}x\]
这题看上去有些像初中数学竞赛里面的因式分解?啊,果然!初中学的东西还是有点用的hhh

就直接写过程咯

\[
\begin{aligned}
原式 &= \int \left(\frac{2 x+1}{x^2+4}+\frac{2}{x^2+1}-\frac{2 x}{\left(x^2+4\right)^2}-\frac{2}{x}\right) \, \text{d}x \\
&= 2 \int \frac{1}{x^2+1} \, \text{d}x+\int \frac{2 x+1}{x^2+4} \, \text{d}x-2 \int \frac{x}{\left(x^2+4\right)^2} \, \text{d}x-2 \int \frac{1}{x} \, \text{d}x \\
&= 2 \int \frac{1}{x^2+1} \, \text{d}x-2 \int \frac{x}{\left(x^2+4\right)^2} \, \text{d}x+\int \left(\frac{2 x}{x^2+4}+\frac{1}{x^2+4}\right) \, \text{d}x-2 \int \frac{1}{x} \, \text{d}x \\
&= 2 \int \frac{1}{x^2+1} \, \text{d}x+\int \frac{1}{x^2+4} \, \text{d}x+2 \int \frac{x}{x^2+4} \, \text{d}x-2 \int \frac{x}{\left(x^2+4\right)^2} \, \text{d}x-2 \int \frac{1}{x} \, \text{d}x \\
\end{aligned}
\]

看到这里有何感想?当然是换元辣!

作换元\( u= x^2+4\)

\[
\begin{aligned}
原式 &= \ln u +2 \int \frac{1}{x^2+1} \, \text{d}x+\int \frac{1}{x^2+4} \, \text{d}x-2 \int \frac{x}{\left(x^2+4\right)^2} \, \text{d}x-2 \int \frac{1}{x} \, \text{d}x \\
&= \ln u + \int \frac{1}{4 \left(\frac{x^2}{4}+1\right)} \, \text{d}x+2 \int \frac{1}{x^2+1} \, \text{d}x-2 \int \frac{x}{\left(x^2+4\right)^2} \, \text{d}x-2 \int \frac{1}{x} \, \text{d}x \\
&= \ln u + \frac{1}{4} \int \frac{1}{\frac{x^2}{4}+1} \, \text{d}x+2 \int \frac{1}{x^2+1} \, \text{d}x-2 \int \frac{x}{\left(x^2+4\right)^2} \, \text{d}x-2 \int \frac{1}{x} \, \text{d}x \\
\end{aligned}
\]

然后继续作换元!想不到吧!哈哈哈哈

作换元\( s = \dfrac{x}{2}, p = x^2 + 4\) (为了确保积分的严谨性,每个单独的积分采用一个单独的变量换元)

\[
\begin{aligned}
原式 &= \frac{1}{2}\arctan (s) + \ln (u) – \int {\frac{1}{{{p^2}}}} {\mkern 1mu} {\text{d}}p + 2\int {\frac{1}{{{x^2} + 1}}} {\mkern 1mu} {\text{d}}x – 2\int {\frac{1}{x}} {\mkern 1mu} {\text{d}}x \\
&= \frac{1}{p} + \frac{1}{2}\arctan (s) + \ln u – 2\ln x + 2\arctan (x) + C \\
&= \frac{1}{{{x^2} + 4}} + \ln \left( {{x^2} + 4} \right) – 2\ln \left( x \right) + \frac{1}{2}\arctan \left( {\frac{x}{2}} \right) + 2\arctan \left( x \right)
\end{aligned}
\]

喵的累死…(酱紫君:

最后留一道作业题吧
\[ \int {{{\sin }^3}(2x){{\cos }^2}(3x){\text{d}}x} \]

这题嘛…仔细想想三角变换…不要上来就换元…(我来想想你可能换些啥呢)

打赏作者
0

1 对 “一些不是很有趣的积分水题(1)”的想法;

  1. 答案:
    $$
    \frac{1}{{384}}( – 72\cos (2x) + 18\cos (4x) + 8\cos (6x) – 9\cos (8x) + 2\cos (12x)) + C
    $$
    有巧妙的过程的可以发,LaTeX公式用两个美元的那个玩意儿扣起来就行

    0

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