简单不定积分一道∫[x+(1-x^2)^(1/2)]^(-1)就这样.

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/19 03:26:10
简单不定积分一道∫[x+(1-x^2)^(1/2)]^(-1)就这样.
x){i';z|mOv4lcutnEffl _lv="}8 "_`gC[^Ra[WRa_\Rs>ЈGUEA5@ZAL͔]]]G CMfG A 6HXSDb1ԍ !KeJtr0jk7VMmv#)AimN_ 5yo:D6b07ֶDEܧ3A&JV'%+*RQ[od_\gJW-

简单不定积分一道∫[x+(1-x^2)^(1/2)]^(-1)就这样.
简单不定积分一道
∫[x+(1-x^2)^(1/2)]^(-1)
就这样.

简单不定积分一道∫[x+(1-x^2)^(1/2)]^(-1)就这样.
令x=sint,dx=costdt
∫1/[x+√(1-x^2)]dx=∫cost/(sint+cost)dt---A
(1)
A=∫cost/(sint+cost)dt=∫[(cost+sint)-sint]/(sint+cost)dt=∫1-[sint/(sint+cost)]dt=t-∫sint/(sint+cost)dt
(2)
A=∫cost/(sint+cost)dt=∫[(cost-sint)+sint]/(sint+cost)dt=∫(cost-sint)/(sint+cost)dt+∫[sint/(sint+cost)dt=ln(sint+cost)+∫sint/(sint+cost)dt
(1)+(2),得
2A=[t-∫sint/(sint+cost)dt]+[ln(sint+cost)+∫sint/(sint+cost)dt]=t+ln(sint+cost)
则A=[t+ln(sint+cost)]/2={arc sinx+ln[x+√(1-x^2)]}/2