原式=∫[∫[-π/2,π/2][cosxdx/(1+(cosx)^2]+∫[-π/2,π/2][xcosxdx/(1+(cosx)^2]
前一项是偶函数,后一项是奇函数,积分为0,
原式=2∫[0,π/2][cosxdx/(1+(cosx)^2]
=2∫[0,π/2]dsinx/[2-(sinx)^2]
设t=sinx,
原式=2∫[0,1]]dt/(2-t^2)
=2*/(2√2）ln|√2-t|/|√2+t|[0,1]
=（√2/2）ln(3+2√2).
求定积分∫上限π/2下限-π/2 (1+x)cosx/1+cos^2xdx
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