Question

不定积分高数∫x/√1+x-x^2 dx

Answer 1

原式 =∫x/√[5/4-(x-1/2)^2]dx
先积∫1/√[(√5/2)^2-(x-1/2)^2]d(x-1/2)=∫1/√[(√5/2)^2-t^2]dt 《t=x-1/2》
∫1/√[(√5/2)^2-t^2]dt=arcsint/(√5/2)+C=2√5/5arcsin(x-1/2)+C
原式=2√5/5∫xd[arcsin(x-1/2)]=2√5/5xarcsin(x-1/2)-2√5/5∫[arcsin(x-1/2)]dx
现在积∫[arcsin(x-1/2)]d(x-1/2)=∫arcsinhdh=harcsinh-∫hdarcsinh
=harcsinh-∫h/√(1-h^2)dx
=harcsinh+√(1-h^2)+C=(x-1/2)arcsin(x-1/2)+√[1-(x-1/2)^2]+C
原式=2√5/5xarcsin(x-1/2)-2√5/5{(x-1/2)arcsin(x-1/2)+√[1-(x-1/2)^2]}+C

Answer 2