求解一道高数定积分题目,谢谢
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设 u = √(e^x - 1),则 x = ln(u²+1),dx = 2udu/(u²+1),u = 1 → √3
那么,原胡旁悉积分就裤乎变换为:
=∫启早2udu/[(u²+1) * u]
=∫2du/(u²+1)
=2∫du/(u²+1)
=2arctan(u)|u = 1 → √3
=2 * [arctan(√3) - arctan(1)]
=2 * (π/3 - π/4)
=π/6
那么,原胡旁悉积分就裤乎变换为:
=∫启早2udu/[(u²+1) * u]
=∫2du/(u²+1)
=2∫du/(u²+1)
=2arctan(u)|u = 1 → √3
=2 * [arctan(√3) - arctan(1)]
=2 * (π/3 - π/4)
=π/6
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