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原式=∫(0,1)(x+1)d(x+1)/(x^2+2x+3)
=1/2∫(0,1)d(x+1)^2/(x^2+2x+3)
=1/2∫(0,1)d(x^2+2x+1)/(x^2+2x+3)
=1/2∫(0,1)d(x^2+2x+3)/(x^2+2x+3)
=ln(x^2+2x+3)/2|(0,1)
=(ln6-ln3)/2
=ln2/2
=1/2∫(0,1)d(x+1)^2/(x^2+2x+3)
=1/2∫(0,1)d(x^2+2x+1)/(x^2+2x+3)
=1/2∫(0,1)d(x^2+2x+3)/(x^2+2x+3)
=ln(x^2+2x+3)/2|(0,1)
=(ln6-ln3)/2
=ln2/2
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