求一道不定积分,谢谢了.dx/x8次方(1-x2次方),多谢了.

显示全部楼层 · 2019-6-25 15:33:28

dx/x8次方(1-x2次方)的不定积分结果为-1/(7*x^7)-1/(5*x^5)-1/(3*x^3)-1/x+1/2ln|(x+1)/(x-1)|+C。解：∫1/(x^8*(1-x^2))dx=∫(1-x^2+x^2)/(x^8*(1-x^2))dx=∫(1-x^2)/(x^8*(1-x^2))dx+∫x^2/(x^8*(1-x^2))dx=∫1/x^8dx+∫1/(x^6*(1-x^2))dx=-1/(7x^7)+∫(1-x^2+x^2)/(x^6*(1-x^2))dx=-1/(7x^7)+∫1/x^6dx+∫1/(x^4*(1-x^2))dx=-1/(7*x^7)-1/(5*x^5)+∫(1-x^2+x^2)/(x^4*(1-x...

千问 · 2019-6-25 15:33:28

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