Question

求不定积分∫x/(x^2+2x+2)dx

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Answer 1

答：
∫[x/(x^2+2x+2)]dx

=∫ {(x+1-1)/[(x+1)²+1]}dx

=∫{(x+1)/[(x+1)²+1]}d(x+1) - ∫ {1/[(x+1)²+1]} d(x+1)

=(1/2)∫ {1/[(x+1)²+1]} d[(x+1)²+1] -  ∫ {1/[(x+1)²+1]} d(x+1)

=(1/2) ln [(x+1)²+1] -arctan(x+1)+C

= ln√(x²+2x+2) -arctan(x+1)+C

Answer 2

解
∫x/(x²+2x+2)dx
=1/2∫(2x+2-2)/(x²+2x+2)dx
=1/2∫(2x+2)/(x²+2x+2)dx-∫1/(x²+2x+2)dx
=1/2∫1/(x²+2x+2)d(x²+2x+2)-∫1/[(x+1)²+1]dx
=1/2∫1/udu-∫1/[(x+1)²+1]d(x+1)
=1/2ln|u|-∫1/(u²+1)du
=1/2ln(x²+2x+2)-acrtanu+C
=1/2ln(x²+2x+2)-arctan(x+1)+C