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?求不

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