dx/(x^2 - 4x)的不定积分

Question

求方法，步骤。谢

Answer 1

原式=∫dx/(x^2-4x+4-4)
=∫d(x-2)/[(x-2)^2-4] (变为导（x-2），式子不变)
令x-2=2sect,则x=2sect+2,故dx=2sect*tantdt
所以原式=∫dt/[4((sect)^2-1] *2sect*tant
=∫dt/(2tant) *sect
=1/2*∫dt/sint
=1/2*∫dt*sint/(sint)^2 (上下同时乘一个sint)
=1/2*∫dcost/[1-(cost)^2] (将sint放入d里面)
=1/4*[∫dcost/(1-cost) +∫dcost/(1+cost)] (将分母裂项，同时前面提个1/2出来)
=1/4*[-ln(1-t)+ln(1+t)+c]
=ln[(1+t)/(1-t)]+c
=ln[(1+arccos(2/(x-2)))/(1-arccos(2/(x-2)))+c
(将前面2sect=x-2化为用x表示t的形式即完成了。)

Answer 2

1/3x^3-4*1/2x^2=1/3x^3-2x^2

Answer 3

∫dx/(x^2 - 4x)
=∫(1/4)(1/(x-4)-1/x)dx
=(1/4)∫dx/(x-4)-(1/4)∫dx/x
=(1/4)ln|x-4|-(1/4)ln|x|+C
=(ln|(x-4)/x|)/4+C本回答被提问者采纳