∫1/√(x^2+2x+2)dx?

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∫1/√(x^2+2x+2)dx?
∫1/√(x^2+2x+2)dx?

∫1/√(x^2+2x+2)dx?
∫1/√(x^2+2x+2) dx
=∫1/√[(x^2+1)^2+1] dx,令u=x+1
=∫1/√(u^2+1) du,令u=tany,du=sec^2y dy
=∫sec^2y/secy dy
=ln(secy+tany)+C
=ln[u+√(u^2+1)]+C
=ln[(x+1)+√(x^2+2x+2)]+C
或=arcsinh(x+1)+C

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