我想你的题应该是∫1/(x+√(1-x²))dx吧?
令x=sinu，√(1-x²)=cosu，dx=cosudu
∫1/(x+√(1-x²))dx
=∫1/(sinu+cosu)*(cosu)du
=∫cosu/(sinu+cosu)du
=1/2∫(cosu+sinu+cosu-sinu)/(sinu+cosu)du
=1/2∫(cosu+sinu)/(sinu+cosu)du+1/2∫(cosu-sinu)/(sinu+cosu)du
=1/2∫1du+1/2∫1/(sinu+cosu)d(sinu+cosu)
=(1/2)u+(1/2)ln|sinu+cosu|+C
=(1/2)arcsinx+(1/2)ln|x+√(1-x²)|+C
令x=sinu，√(1-x²)=cosu，dx=cosudu
∫1/(x+√(1-x²))dx
=∫1/(sinu+cosu)*(cosu)du
=∫cosu/(sinu+cosu)du
=1/2∫(cosu+sinu+cosu-sinu)/(sinu+cosu)du
=1/2∫(cosu+sinu)/(sinu+cosu)du+1/2∫(cosu-sinu)/(sinu+cosu)du
=1/2∫1du+1/2∫1/(sinu+cosu)d(sinu+cosu)
=(1/2)u+(1/2)ln|sinu+cosu|+C
=(1/2)arcsinx+(1/2)ln|x+√(1-x²)|+C
温馨提示：内容为网友见解，仅供参考
第1个回答 2021-02-01
令a=1即可，原式=
(1/2)arcsinx+(1/2)ln|x+√(1-x²)|+C
第2个回答 2012-04-09
=lnx+1/2arcsinx+1/2x√(1-x^2)+C
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