求不定积分∫[(2x+1)/(x*x-2x+2)]dx?要有详细的解答哦..谢谢了..∫[(2x-2)/(x^2-2x+2)]dx 到∫[1/(x^2-2x+2)]d(x^2-2x+2)是为什么啊?
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![求不定积分∫[(2x+1)/(x*x-2x+2)]dx?要有详细的解答哦..谢谢了..∫[(2x-2)/(x^2-2x+2)]dx 到∫[1/(x^2-2x+2)]d(x^2-2x+2)是为什么啊?](/uploads/image/z/1633018-58-8.jpg?t=%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%AB%5B%EF%BC%882x%2B1%EF%BC%89%2F%28x%2Ax-2x%2B2%29%5Ddx%3F%E8%A6%81%E6%9C%89%E8%AF%A6%E7%BB%86%E7%9A%84%E8%A7%A3%E7%AD%94%E5%93%A6..%E8%B0%A2%E8%B0%A2%E4%BA%86..%E2%88%AB%5B%EF%BC%882x-2%EF%BC%89%2F%28x%5E2-2x%2B2%29%5Ddx+++%E5%88%B0%E2%88%AB%5B1%2F%28x%5E2-2x%2B2%29%5Dd%28x%5E2-2x%2B2%EF%BC%89%E6%98%AF%E4%B8%BA%E4%BB%80%E4%B9%88%E5%95%8A%EF%BC%9F)
求不定积分∫[(2x+1)/(x*x-2x+2)]dx?要有详细的解答哦..谢谢了..∫[(2x-2)/(x^2-2x+2)]dx 到∫[1/(x^2-2x+2)]d(x^2-2x+2)是为什么啊?
求不定积分∫[(2x+1)/(x*x-2x+2)]dx?
要有详细的解答哦..谢谢了..
∫[(2x-2)/(x^2-2x+2)]dx 到∫[1/(x^2-2x+2)]d(x^2-2x+2)是为什么啊?
求不定积分∫[(2x+1)/(x*x-2x+2)]dx?要有详细的解答哦..谢谢了..∫[(2x-2)/(x^2-2x+2)]dx 到∫[1/(x^2-2x+2)]d(x^2-2x+2)是为什么啊?
原式=∫[(2x-2+3)/(x^2-2x+2)]dx
=∫[(2x-2)/(x^2-2x+2)]dx+∫[3/(x^2-2x+2)]dx
=∫[1/(x^2-2x+2)]d(x^2-2x+2)+3∫{1/[(x-1)^2+1]}d(x-1)
=ln(x^2-2x+2)+3arctan(x-1)+C
楼主所说的∫[(2x-2)/(x^2-2x+2)]dx 到∫[1/(x^2-2x+2)]d(x^2-2x+2)
其实就是典型的凑微分方法 因为(2x-2)dx=d(x^2-2x)=d(x^2-2x+2)
这种很明显要用凑微分的方法嘛
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