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原式=∫[0,(√3+1)]xdx/[(x-1)²+1]
令x-1=tant，则dx=sec²t dt
x=0时，t=-π/4，x=√3+1时，t=π/3.
故原式=∫[-π/4,π/3] (1+tant)sec²tdt /sec²t
=∫[-π/4,π/3] (1+tant)dt
=(t - ln|cost|)|[-π/4,π/3]
=(π/3-ln½) - [-π/4-ln(√2/2)]
=π/3 + ln2 +π/4 +ln(1/√2)
=7π/12+ln2-ln√2
=7π/12+ln2-½ ln2
=7π/12 +½ln2
令x-1=tant，则dx=sec²t dt
x=0时，t=-π/4，x=√3+1时，t=π/3.
故原式=∫[-π/4,π/3] (1+tant)sec²tdt /sec²t
=∫[-π/4,π/3] (1+tant)dt
=(t - ln|cost|)|[-π/4,π/3]
=(π/3-ln½) - [-π/4-ln(√2/2)]
=π/3 + ln2 +π/4 +ln(1/√2)
=7π/12+ln2-ln√2
=7π/12+ln2-½ ln2
=7π/12 +½ln2
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