∫(x+√(1-x^2))dx
=∫xdx+∫√(1-x^2)dx
∫√(1-x^2)dx 取x=sinu =∫cosu^2du=(u/2)+(sinucosu)/2 =(1/2)arcsinx+(1/2)x√(1-x^2)
=x^2/2+(1/2)arcsinx+(1/2)x√(1-x^2)+C
∫(x+√(1-x^2))dx
=∫xdx+∫√(1-x^2)dx
∫√(1-x^2)dx 取x=sinu =∫cosu^2du=(u/2)+(sinucosu)/2 =(1/2)arcsinx+(1/2)x√(1-x^2)
=x^2/2+(1/2)arcsinx+(1/2)x√(1-x^2)+C