∫01arctanx/(1+x2)2dx=________.

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问题01arctanx/(1+x2)2dx=________.

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答案π2/64+π/16-1/8

解析01arctanx/(1+x2)2dx→∫0π/4t/sec4t·sec2tdt=∫0π/4tcos2tdt=1/2∫0π/4t(1+cos2t)dt=π2/64+1/8∫0π/42tcos2td(2t)-π2/64+1/8∫0π/2tcostdt=π2/64+1/8(tsint|0π/2-∫0π/4sintdt)=π2/64+1/8(π/2-1)=π2/64+π/16-1/8.
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