瞎做积分题笔记

瞎找题系列1

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花了半个下午+半个晚上做完了，发现其实一直做脑筋急转弯就好了。
这里记录点转了蛮久弯才做出来的题吧。

首先来自己感觉蛮有用的:

• 遇式不决，分式分解。

T35

\[\begin{aligned} \int&\frac{1-\ln x}{(x-\ln x)^2}\ dx\\ =\int&\frac{1}{(1 - \frac {\ln x}x)^2}\frac{1-\ln x}{x^2}\ dx\\ \begin{equation} \xlongequal{t=\frac{\ln x}x - 1} \end{equation}&\int\frac1{t^2}\ dt\\ =&-\frac1t+C\\ =&\frac{x}{x-\ln x}+C \end{aligned} \]

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T58

\[\begin{aligned} &\int\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}\ dx\\ =&\int1-\frac2{\sqrt{x+1}+1}\ dx\\ \begin{equation} \xlongequal{t=\sqrt{x+1}+1} \end{equation}&x-\int\frac2t d(t^2-2t)+C\\ =&x-\int\frac{4t-4}t\ dt+C\\ =&x-4\sqrt{x+1}+4\ln(\sqrt{x+1}+1)+C \end{aligned} \]

T68

\[\begin{aligned} &\int\frac{\sin x}{\sin x+\cos x}\ dx\\ =&\int1-\frac{1}{1+\tan x}\ dx\\ \begin{equation}\xlongequal{t=\tan x}\end{equation}&x-\int\frac1{1+t}\ d\arctan t+C\\ =&x-\int\frac1{(1+t)(1+t^2)}\ dt+C\\ =&x-\frac12\int(\frac1{1+t}-\frac{t-1}{1+t^2})\ dt+C\\ =&x-\frac12\ln|1+\tan x|+\frac14\ln(\tan^2x+1)-\frac12x+C\\ =&\frac12(x-\ln|1+\tan x|+\frac12\ln(\tan^2x+1))+C\\ \end{aligned} \]