记录入群题的解答

作者: 洛玖言 | 来源:发表于2019-11-13 18:31 被阅读0次

记录入群题的解答

只是提供一种解法，并不是最优的，为了清楚，有的写的比较繁琐.

1

$I=\int_0^1\dfrac{\ln(1+x)}{1+x^2}\text{d} x$

Sol:
$\begin{aligned} x=&\tan t\\ I=&\int_0^{\frac\pi4}\ln(1+\tan t)\text{d}t\\ =&\int_0^{\frac\pi4}\ln\left[1+\tan\left(\dfrac{\pi}{4}-x\right)\right]\text{d}x\\ =&\int_0^{\frac\pi4}\ln\left(1+\dfrac{1-\tan x}{1+\tan x}\right)\text{d}x\\ =&\int_0^{\frac\pi4}\ln\left(\dfrac{2}{1+\tan x}\right)\text{d}x\\ =&\int_0^{\frac\pi4}\ln2\text{d}x-\int_0^{\frac\pi4}\ln(1+\tan x)\text{d}x\\ =&\dfrac{\pi}{4}\ln2-I\\ 2I=&\dfrac{\pi}{4}\ln2\\ I=&\dfrac{\pi}{8}\ln2 \end{aligned}$

2

$I=\lim_{n\to\infty}\dfrac{n}{\sqrt[n]{n!}}$

Sol:
$\begin{aligned} \dfrac{1}{I}=&\lim_{n\to\infty}\dfrac{\sqrt[n]{n!}}{n}=\lim_{ n\to\infty}\sqrt[n]{\dfrac{n!}{n^n}}=\lim_{n\to\infty}\sqrt[n]{\dfrac{1\cdot 2\cdot3\cdots n}{n\cdot n\cdot n\cdots n}}\\ =&\lim_{n\to\infty}\exp\left[{\frac1n(\ln\frac1n+\frac2n+\frac3n+\cdots+\frac nn)}\right]\\ =&\lim_{n\to\infty}\exp\left[\int_0^1\ln x\text{d}x\right]\\ =&\exp\left[x\ln x\bigg|_0^1-\int_0^1\text{d}x\right]\\ =&\frac{1}{e}\\ I=&e \end{aligned}$

3

$\int_0^1\dfrac{ \arcsin\sqrt{x}}{\sqrt{1-x+x^2}}\text{d}x$

Sol:
$\begin{aligned} &令\;t=1-x\\ I=&\int_0^1\dfrac{\arcsin\sqrt{1-t}}{\sqrt{1-t+t^2}}\text{d}t\\ =&\dfrac12\left[\int_0^1\dfrac{\arcsin\sqrt{x}}{\sqrt{1-x+x^2}}\text{d}x+\int_0^1\dfrac{\arcsin\sqrt{1-x}}{\sqrt{1-x+x^2}}\text{d}x\right]\\ =&\dfrac12\int_0^1\dfrac{\arcsin\sqrt{x}+\arcsin\sqrt{1-x}}{\sqrt{1-x+x^2}}\text{d}x\\ =&\dfrac{\pi}{4}\int_0^1\dfrac{\text{d}x}{\sqrt{1-x+x^2}}\\ =&\dfrac\pi4\int_0^1\dfrac{\text{d}x}{\sqrt{(x-\frac12)^2+\frac34}}=\dfrac\pi4\int_{-\frac{1}{\sqrt{3}}}^{\frac1{\sqrt{3}}}\dfrac{\text{d}u}{\sqrt{u^2+1}}\\ =&\dfrac{\pi}{4}\ln3 \end{aligned}$

$\begin{aligned} &x=\tan t\\ &\int\dfrac{\text{d}x}{\sqrt{x^2+1}}=\int\frac{1}{\cos t}\text{d}t=\int\dfrac{\cos t}{1-\sin^2t}\text{d}t\\ =&\int\dfrac{\text{d}\sin t}{1-\sin^2t}\\ =&\dfrac12\int\left(\dfrac{1}{1-\sin t}+\dfrac{1}{1+\sin t}\right)\text{d}\sin t\\ =&\dfrac12\ln\left|\dfrac{1+\sin t}{1-\sin t}\right|+C\\ =&\dfrac12\ln\left|\dfrac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}-x}\right|+C\\ =&\ln|\sqrt{1+x^2}+x|+C=\text{arcsinh}x+C \end{aligned}$

记录入群题的解答

记录入群题的解答

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