howshineyou解法比较巧妙，一般人很难想到，下面是比较一般做法，供参考。
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第1个回答 2013-12-31
原式是偶函数
所以
原式=2∫(0,+∞)(x²＋1)/(x^4+1)dx
=2∫(0,+∞)(1＋1/x²)/(x²+1/x²)dx
=2∫(0,+∞)1/(x²+1/x²) d(x-1/x)
=2∫(0,+∞)1/(x²+1/x²-2+2) d(x-1/x)
=2∫(0,+∞)1/[(x-1/x)²＋2] d(x-1/x)
=2/√2 arctan(x-1/x)/√2 |（0,+∞）
=√2 (π/2-(-π/2))
=√2 π本回答被提问者采纳
所以
原式=2∫(0,+∞)(x²＋1)/(x^4+1)dx
=2∫(0,+∞)(1＋1/x²)/(x²+1/x²)dx
=2∫(0,+∞)1/(x²+1/x²) d(x-1/x)
=2∫(0,+∞)1/(x²+1/x²-2+2) d(x-1/x)
=2∫(0,+∞)1/[(x-1/x)²＋2] d(x-1/x)
=2/√2 arctan(x-1/x)/√2 |（0,+∞）
=√2 (π/2-(-π/2))
=√2 π本回答被提问者采纳
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