求定积分∫(0,1) 1/(x^2+2x+2)dx

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求定积分∫(0,1) 1/(x^2+2x+2)dx
求定积分∫(0,1) 1/(x^2+2x+2)dx

求定积分∫(0,1) 1/(x^2+2x+2)dx
原式=∫(0→1)dx/((x+1)^2+1)
=∫(0→1)d(x+1)/((x+1)^2+1)
=arctan(x+1)|(0→1)
=arctan2-π/4

原式=∫(0,1) 1/[1+(x+1)²]d(x+1)
=arctan(x+1) (0,1)
=arctan2-π/4

原式=∫(0→1)dx/((x+1)^2+1)
=∫(0→1)d(x+1)/((x+1)^2+1)
=arctan(x+1)|(0→1)
=arctan2-π/4