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
[egin{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\
egin{equation}
xlongequal{t=frac{ln x}x - 1}
end{equation}&intfrac1{t^2} dt\
=&-frac1t+C\
=&frac{x}{x-ln x}+C
end{aligned}
]
T58
[egin{aligned}
&intfrac{sqrt{x+1}-1}{sqrt{x+1}+1} dx\
=&int1-frac2{sqrt{x+1}+1} dx\
egin{equation}
xlongequal{t=sqrt{x+1}+1}
end{equation}&x-intfrac2t d(t^2-2t)+C\
=&x-intfrac{4t-4}t dt+C\
=&x-4sqrt{x+1}+4ln(sqrt{x+1}+1)+C
end{aligned}
]
T68
[egin{aligned}
&intfrac{sin x}{sin x+cos x} dx\
=&int1-frac{1}{1+ an x} dx\
egin{equation}xlongequal{t= an x}end{equation}&x-intfrac1{1+t} darctan t+C\
=&x-intfrac1{(1+t)(1+t^2)} dt+C\
=&x-frac12int(frac1{1+t}-frac{t-1}{1+t^2}) dt+C\
=&x-frac12ln|1+ an x|+frac14ln( an^2x+1)-frac12x+C\
=½(x-ln|1+ an x|+frac12ln( an^2x+1))+C\
end{aligned}
]