第二题求解答，谢谢
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(2)
由积分区间，令x=sint
x：-½→½，则t：-π/6→π/6
∫[-½：½] [(x+1)/√(1-x²)]dx
=∫[-π/6：π/6] [(sint +1)/√(1-sin²t)]d(sint)
=∫[-π/6：π/6] [(sint +1)cost/cost]dt
=∫[-π/6：π/6] (sint +1)dt
=(-cost +t)|[-π/6：π/6]
=[-cos(π/6)+ π/6]-[-cos(-π/6)+(-π/6)]
=-cos(π/6)+π/6 +cos(π/6)+π/6
=π/3
由积分区间，令x=sint
x：-½→½，则t：-π/6→π/6
∫[-½：½] [(x+1)/√(1-x²)]dx
=∫[-π/6：π/6] [(sint +1)/√(1-sin²t)]d(sint)
=∫[-π/6：π/6] [(sint +1)cost/cost]dt
=∫[-π/6：π/6] (sint +1)dt
=(-cost +t)|[-π/6：π/6]
=[-cos(π/6)+ π/6]-[-cos(-π/6)+(-π/6)]
=-cos(π/6)+π/6 +cos(π/6)+π/6
=π/3
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