不定积分换元法∫(x/1+x^2)dx=1/2∫(dx^2/1+x^2)=1/2∫(du/1+u)=1/2∫[d(u+1)/1+u]我想问的是∫(x/1+x^2)dx=1/2∫(dx^2/1+x^2)这一步怎么计算出来的,还有为什么1/2∫(du/1+u)=1/2∫[d(u+1)/1+u]中的du=d(u+1)?
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![不定积分换元法∫(x/1+x^2)dx=1/2∫(dx^2/1+x^2)=1/2∫(du/1+u)=1/2∫[d(u+1)/1+u]我想问的是∫(x/1+x^2)dx=1/2∫(dx^2/1+x^2)这一步怎么计算出来的,还有为什么1/2∫(du/1+u)=1/2∫[d(u+1)/1+u]中的du=d(u+1)?](/uploads/image/z/6946243-43-3.jpg?t=%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E6%8D%A2%E5%85%83%E6%B3%95%E2%88%AB%28x%2F1%2Bx%5E2%29dx%3D1%2F2%E2%88%AB%28dx%5E2%2F1%2Bx%5E2%29%3D1%2F2%E2%88%AB%28du%2F1%2Bu%29%3D1%2F2%E2%88%AB%5Bd%28u%2B1%29%2F1%2Bu%5D%E6%88%91%E6%83%B3%E9%97%AE%E7%9A%84%E6%98%AF%E2%88%AB%28x%2F1%2Bx%5E2%29dx%3D1%2F2%E2%88%AB%28dx%5E2%2F1%2Bx%5E2%29%E8%BF%99%E4%B8%80%E6%AD%A5%E6%80%8E%E4%B9%88%E8%AE%A1%E7%AE%97%E5%87%BA%E6%9D%A5%E7%9A%84%2C%E8%BF%98%E6%9C%89%E4%B8%BA%E4%BB%80%E4%B9%881%2F2%E2%88%AB%28du%2F1%2Bu%29%3D1%2F2%E2%88%AB%5Bd%28u%2B1%29%2F1%2Bu%5D%E4%B8%AD%E7%9A%84du%3Dd%28u%2B1%29%3F)
不定积分换元法∫(x/1+x^2)dx=1/2∫(dx^2/1+x^2)=1/2∫(du/1+u)=1/2∫[d(u+1)/1+u]我想问的是∫(x/1+x^2)dx=1/2∫(dx^2/1+x^2)这一步怎么计算出来的,还有为什么1/2∫(du/1+u)=1/2∫[d(u+1)/1+u]中的du=d(u+1)?
不定积分换元法
∫(x/1+x^2)dx=1/2∫(dx^2/1+x^2)=1/2∫(du/1+u)=1/2∫[d(u+1)/1+u]
我想问的是∫(x/1+x^2)dx=1/2∫(dx^2/1+x^2)这一步怎么计算出来的,还有为什么1/2∫(du/1+u)=1/2∫[d(u+1)/1+u]中的du=d(u+1)?
不定积分换元法∫(x/1+x^2)dx=1/2∫(dx^2/1+x^2)=1/2∫(du/1+u)=1/2∫[d(u+1)/1+u]我想问的是∫(x/1+x^2)dx=1/2∫(dx^2/1+x^2)这一步怎么计算出来的,还有为什么1/2∫(du/1+u)=1/2∫[d(u+1)/1+u]中的du=d(u+1)?
首先你要懂得导数的运算公式,求不定积分是求导的逆过程
∫ x/(1 + x²) dx
= ∫ 1/(1 + x²) • (x dx)
= ∫ 1/(1 + x²) d(x²/2)
这里其实是对x求积分的,即x dx ∫ x dx = x²/2 + C d(x²/2 + C) = d(x²/2),C在求导后变为0
或者用导数容易理解,就是(x²/2)' = d(x²/2)/dx = 1/2 • 2x = x
变为微分形式就是d(x²/2) = x dx
再次根据求导的原理
由于任何常数的导数都是0
d(C)/dx = 0 ==> d(C) = 0
而d(Cx)/dx = C • dx/dx = C ==> d(Cx) = C • dx
再进一步,d(Ax + B)/dx = (A + 0) = A ==> d(Ax + B) = A • dx
于是d(u + 1)/du = (u + 1)' = 1,u + 1对u求导
得出d(u + 1) = du,两边乘以du即可,这是微分形式
0.5dx^2=0.5*(x^2)’dx=0.5*2xdx=xdx
因为(u+1)’=u‘=u‘+1’=u‘+0=u‘
所以du=d(u+1)