∫[x/√(x^2+2x+2)]dx
=∫{[(x+1)-1]/√[(x+1)^2+1]}dx
=∫{(x+1)/√[(x+1)^2+1]}dx-∫{1/√[(x+1)^2+1]}dx
=(1/2)∫{1/√[(x+1)^2+1]}d[(x+1)^2+1]-∫{1/√[(x+1)^2+1]}dx
=∫d{√[(x+1)^2+1]}-∫{1/√[(x+1)^2+1]}dx
=√[(x+1)^2+1]-∫{1/√[(x+1)^2+1]}dx
=√(x^2+2x+2)-∫{1/√[(x+1)^2+1]}dx.
令x+1=tanu,则:dx=[1/(cosu)^2]du.
∴∫[x/√(x^2+2x+2)]dx
=√(x^2+2x+2)-∫{1/√[(tanu)^2+1]}[1/(cosu)^2]du
=√(x^2+2x+2)-∫[cosu/(cosu)^2]du
=√(x^2+2x+2)-∫{1/[(1+sinu)(1-sinu)]}d(sinu)
=√(x^2+2x+2)-(1/2)∫[1/(1+sinu)+1/(1-sinu)]d(sinu)
=√(x^2+2x+2)-(1/2)∫[1/(1+sinu)]d(sinu)-(1/2)∫[1/(1-sinu)]d(sinu)
=√(x^2+2x+2)-(1/2)ln(1+sinu)+(1/2)ln(1-sinu)+C
=√(x^2+2x+2)+(1/2)ln{1+tanu/√[(tanu)^2+1]}
-(1/2)ln{1-tanu/√[(tanu)^2+1]}+C
=√(x^2+2x+2)+(1/2)ln{√[(tanu)^2+1]+tanu}
-(1/2)ln{√[(tanu)^2+1]-tanu}+C
=√(x^2+2x+2)+(1/2)ln{√[(x+1)^2+1]+x+1}
-(1/2)ln{√[(x+1)^2+1]-x-1}+C
=√(x^2+2x+2)+(1/2)ln{[√(x^2+2x+2)+x+1]^2}+C
=√(x^2+2x+2)+ln[√(x^2+2x+2)+x+1]+C.
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