求∫1/（x²-9）dx

更新时间2019-04-08 10:29:12

∫1/（x²-9）dx

=∫1/[(x+3)(x-3)]dx

=1/6*∫6/[(x+3)(x-3)]dx

=1/6*∫[1/(x-3)-1/(x+3)]dx

=1/6*[∫1/(x-3)dx-∫1/(x+3)dx]

=1/6*(ln│x-3│-ln│x+3│)+C

∫[1/(x²-9)]dx

=(1/6)∫[1/(x-3)-1/(x+3)]dx

=(1/6)(ln|x-3|-ln|x+3|)+c

∫1/（x²-9）dx

=∫dx/(x+3)(x-3)

处理到这里，1/(x+3)(x-3)可以化为以下形式

A/(x+3)+B/(x-3)

将上式通分并合并得

(B(x+3)+A(x-3))/(x+3)(x-3)

让分子等于1

即(B(x+3)+A(x-3))=1

Bx+3B+Ax-3A=1

（A+B）x+3(B-A)=1

等式右边没有x因此A+B=0❶

等式右边只有1因此3(B-A)=1或B-A=1/3❷

❶+❷得

2B=1/3或B=1/6

所以A=-1/6

接着

∫dx/(x+3)(x-3)

∫(-1/6)/(x+3)+(1/6)/(x-3)dx

-In|x+3|/6+In|x-3|/6+C