arctanx/(1+x²)² dx 的定积分 上限是1 下限是0
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令x=tant
则原式=∫(0→π/4)t/(1/cos^2(t))^2*1/cos^2(t)dt
=∫(0→π/4)tcos^2(t)dt
=∫(0→π/4)t*(cos(2t)+1)/2dt
=1/2∫(0→π/4)tcos(2t)dt+1/2∫(0→π/4)tdt
=1/4∫(0→π/4)td(sin(2t))+t^2/4|(0→π/4)
=tsin(2t)/4|(0→π/4)-1/4∫(0→π/4)sin(2t)dt+t^2/4|(0→π/4)
=tsin(2t)/4|(0→π/4)+cos(2t)/8|(0→π/4)+t^2/4|(0→π/4)
=π/16-1/8+π^2/64
则原式=∫(0→π/4)t/(1/cos^2(t))^2*1/cos^2(t)dt
=∫(0→π/4)tcos^2(t)dt
=∫(0→π/4)t*(cos(2t)+1)/2dt
=1/2∫(0→π/4)tcos(2t)dt+1/2∫(0→π/4)tdt
=1/4∫(0→π/4)td(sin(2t))+t^2/4|(0→π/4)
=tsin(2t)/4|(0→π/4)-1/4∫(0→π/4)sin(2t)dt+t^2/4|(0→π/4)
=tsin(2t)/4|(0→π/4)+cos(2t)/8|(0→π/4)+t^2/4|(0→π/4)
=π/16-1/8+π^2/64
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