积分问题3
解:u=x+1,dx=du,x=u-1 ````∫[(u-1)/(u²+2)]du ```=(1/2)∫[2u/(u²+2)]du-∫[1/(u²+2)]du ```=(1/2)ln(u²+2)-√2·arctan(u/√2)/2+C ```=(1/2)ln(x²+2x+3)-√2·arctan[(x+1)/√2]/2+C
∫xdx/(x^2+2x+3) =(1/2)∫[(2x+2)-2]dx/(x^2+2x+3) =(1/2)∫d(x^2+2x+3)/(x^2+2x+3)-∫dx/[(x+1)^2+2] =(1/2)ln(x^2+2x+3)-√2∫d[(x+1)/√2]/{[(x+1)/√2]^2+1} =(1/2)ln(x^2+2x+3)-√2arctan[(x+1)/√2]+C.
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