∫lnsin2xdx(0~π/4) (表示从0到π/4的定积分)=∫ln(2sinx cosx)dx(0~π/4)=π/4*ln2+∫lnsinxdx(0~π/4)+∫lncosxdx(0~π/4)=π/4*ln2+∫lnsinxdx(0~π/4)+∫lnsinxdx(π/4~π/2) (对最后一个积分换元)=π/4*ln2+∫lnsinxdx(0~π/2)=π/4*ln2+2∫lnsin2xdx(0...
Integrate[1/Log[x], x] - Wolfram|Alpha (wolframalpha.com)www.wolframalpha.com/input/?i=Integrate%5B1%2FLog%5Bx%5D%2C+x%5D 对数积分li(x)的定义为∫0x1ln(t)dt，而Li(x)为∫2x1ln(t)dt。对数积分_百度百科baike.baidu.com/item/%E5%AF%B9%E6%95%B0%E7%A7%AF%E5%88%...
=∫ln(2sinx cosx)dx(0~π/4)=π/4*ln2+∫lnsinxdx(0~π/4)+∫lncosxdx(0~π/4)=π/4*ln2+∫lnsinxdx(0~π/4)+∫lnsinxdx(π/4~π/2) (对最后一个积分换元)=π/4*ln2+∫lnsinxdx(0~π/2)=π/4*ln2+2∫lnsin2xdx(0~π/4) (换元)由第一个式子与最后一个式子相等即...
\begin{aligned} a_n=&\frac1\pi\int_{-\pi}^{\pi}\cos nx\ln\cos \frac x2dx\\ =&\frac2\pi\underbrace{\int_{0}^{\pi}\cos nx\ln\cos \frac x2dx}_{x\mapsto 2x}\\ =&\frac4\pi\int_{0}^{\frac \pi2}\cos 2nx\ln\cos xdx\\ =&\frac2{n\pi}\int_{0}^{\frac \pi...
1不定积分∫ln(1+x^2)dx 原式=xln(1+x^2)-∫xd[ln(1+x^2)] 这些步骤怎么变的xln(1+x^2)怎么就到∫外了那个x怎么来的希望能讲一下原理课本上一步就到了看不懂怎么回事=xln(1+x^2)-∫2x^2/(1+x^2)dx=xln(1+x^2)-2∫[1-1/(1+x^2)dx=xln(1+x^2)-2x+2acrtgx+C ...
我把具体过程写出来了：
这是Raabe积分：R(0)=∫01lnΓ(x)dx=∫01ln(Γ(x+1)x)dx=∫01lnΓ(x+1)−ln...
【答案】：=xln(1+x2)-2x+2arctanx+C
ln(1+2x)∼2x(x→0)
∫ln(1+x^2)dx (直接分步积分)=xln(1+x^2)-∫x*[ln(1+x^2)]'dx=xln(1+x^2)-∫x*2x/(1+x^2)dx=xln(1+x^2)-2∫(x^2+1-1)/(1+x^2)dx=xln(1+x^2)-2∫[1-1/(1+x^2)]dx=xln(1+x^2)-2∫dx+2∫[1/(1+x^2)]dx=xln(1+x^2)-2x+... 分析总结。 请各位友友...