Question

求不定积分，急~~！

∫[1/x平方(4+x平方)]dx
=1/4∫[1/x平方-1/(4+x平方)]dx

问:1/4是哪里来的???急急

Answer 1

这需要用到待定系数法了
1/[x^2(4+x^2)]
令1/[x^2(4+x^2)]=A/x^2+B/(4+x^2)
通分后比较分子得1=A(4+x^2)+Bx^2
1=4A+Ax^2+Bx^2
1=4A+(A+B)x^2,对比系数得
4A=1得A=1/4
A+B=0得B=-A=-1/4
所以1/[x^2(4+x^2)]=(1/4)(1/x^2)-(1/4)[1/(4+x^2)]=(1/4)[1/x^2-1/(4+x^2)]

原式=(1/4)∫1/x^2 dx-(1/4)∫1/(4+x^2) dx
=1/4*(-1/x)-1/4*1/2*arctan(x/2)+C
=-1/(4x)-(1/8)arctan(x/2)+C