不定積分三角代換 sinsunt

令x=sinu,dx=cosudu∫dx/(x+√(1-x^2))=∫cosudu/(sinu+cosu)=1/2∫(cosu+sinu+cosu-sinu)du/(sinu+cosu)=1/2∫[1+(cosu-sinu)/(sinu+cosu)]du=1/2[u+ln|sinu+cosu|]+C=1/2arcsinx+1/2ln|x+√(1-x^2)|+C