若X＝M是方程(x+3)/6－（x－1）/2＋1＝x－（x－k）/3的解,求｜m＋2n－2｜－｜2n－k｜的值

题目

若X＝M是方程(x+3)/6－（x－1）/2＋1＝x－（x－k）/3的解,求｜m＋2n－2｜－｜2n－k｜的值

答案

(1).看积分式的分子,x=(x+x^(1/2))-(x^(1/2)+1)+(1+x^(-1/2))-x^(-1/2)
所以
∫xdx/1+√x＝∫x^(1/2)dx+∫1dx+∫x^(-1/2)dx+∫[x^(-1/2)/(1+√x)]dx
∫xdx/1+√x＝(2/3)x^(3/2)+x+2x^(1/2)+2*ln(1+x^(1/2))
(2).这个题怎么了?有什么困难?
不是等于√2*x+(1/2)*x^2-(1/3)*x^3吗?
(3).换元,令x=sint
∫dx/1+√1-x^2=∫cost/(1+cost)dt=t-∫1/(1+cost)dt
=t-∫[(1-cost)/(sint)^2]dt
=t+ctan(t)-1/sint
然后把t换回x就行了~!
arcsinx+[√1-(x^2)]/x-1/x