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/etc/nginx/nginx.conf.请问这几道不定积分题怎么做∫dx/x(1-x)∫sinx^8 dx∫x^2 * cos^2x dx∫(1+x)^2 / (1+x^2)能有详细过程最好 感激不尽
∫dx/x(1-x) =∫dx/x-∫d（1-x）/（1-x）=ln|x|-ln|1-x|+C ∫sinx^8 dx（是sinx的8次方还是sin（x的8次方）?） ∫x^2 * cos^2x dx =0.5∫x²（1+cos2x）dx=x³/6+0.5∫x²cos2xdx=x³/6+0.25x²sin2x-0.5∫xsin2xdx=x³/6+0.25x²sin2x-0.25xcos2x+0.5∫cos2xdx=x³/6+0.25x²sin2x-0.25xcos2x+0.25sin2x+C ∫(1+x)^2 / (1+x^2) =∫(1+2x/(1+x²)）dx=x+ln|1+x²| +C
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练多了就会了
∫dx/x(1-x)=∫dx/x+∫dx/(1-x)=∫dx/x-∫d(1-x)/(1-x)=ln|x|-ln|1-x|+C∫sinx^8dx=x*sinx^8-∫xd(sinx^8)=x*sinx^8-8∫x^8*cosx^8dx=x*sinx^8-8∫x^8*(1-sinx^8)dx=x*sinx^8-8∫x^8dx+8∫x^8*sinx^8dx=x*sinx^8-8/9*x^9+8...
扫描下载二维码∫x√(2x+1)dx这道积分怎么做
令√(2x+1)=u,则x=(u²-1)/2,dx=udu∫ x√(2x+1) dx=∫ (u²-1)u²/2 du=(1/2)∫ (u^4-u²) du=(1/10)u^5 - (1/6)u³ + C=(1/10)(2x+1)^(5/2) - (1/6)(2x+1)^(3/2) + C 希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的"选为满意回答"按钮,谢谢.
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换元法去掉根号
扫描下载二维码这道不定积分题怎么做?求过程&
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扫描下载二维码}