df(x)/dx=1/（9+2x+x^2）f(x)=∫dx/（9+2x+x^2）=∫dx/（8+1+2x+x^2）=∫dx/[(x+1)^+8]=(1/8)∫dx/{[(x+1)/2√2]+1}令u=(x+1)/2√2,du=1/2√2dx=(2√2/8)∫du/(u^2+1)=(√2/4)arctanu+C=(√2/4)arctan[(x+1)/2√2]+C...