∫[-1,-3][1/(x^2+4x+5)]dx定积分
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∫[-1,-3][1/(x^2+4x+5)]dx定积分
![∫[-1,-3][1/(x^2+4x+5)]dx定积分](/uploads/image/z/16539267-3-7.jpg?t=%E2%88%AB%5B-1%2C-3%5D%5B1%2F%28x%5E2%2B4x%2B5%29%5Ddx%E5%AE%9A%E7%A7%AF%E5%88%86)
∫[-1,-3][1/(x^2+4x+5)]dx
=∫[-1,-3][1/[(x+2)²+1]d(x+2)
=arctan(x+2)[-1,-3]
是不是-1是上限?
=arctan1-arctan(-1)
=2arctan1
=π/2
再问: 是下限
再答: 哦,那就是相反数
=∫[-1,-3][1/[(x+2)²+1]d(x+2)
=arctan(x+2)[-1,-3]
是不是-1是上限?
=arctan1-arctan(-1)
=2arctan1
=π/2
再问: 是下限
再答: 哦,那就是相反数