求解,这个不定积分? 30
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令1-e^2t=x,则t=½ln(1-x)
dt=-1/[2(1-x)]dx
原式=-¼∫ln(1-x)/x(1-x)dx
=-¼∫ln(1-x)[1/x+1/(1-x)]dx
=-¼∫ln(1-x)/xdx+⅛ln²(1-x)
dt=-1/[2(1-x)]dx
原式=-¼∫ln(1-x)/x(1-x)dx
=-¼∫ln(1-x)[1/x+1/(1-x)]dx
=-¼∫ln(1-x)/xdx+⅛ln²(1-x)
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题目对吗?请附原题印刷版图片。
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Expand the following:
(x (x - 2 log(1 - e^x)))/2 - Li_2(e^x)
x (x - 2 log(1 - e^x)) = x x + x (-2 log(1 - e^x)):
(x x - 2 x log(1 - e^x))/2 - Li_2(e^x)
x x = x^2:
(x^2 - 2 x log(1 - e^x))/2 - Li_2(e^x)
(x^2 - 2 x log(1 - e^x))/2 = x^2/2 - (2 x log(1 - e^x))/2:
(x^2)/2 - (2 x log(1 - e^x))/2 - Li_2(e^x)
(-2)/2 = (2 (-1))/2 = -1:
Answer: |
| x^2/2 + -1 x log(1 - e^x) - Li_2(e^x)
(x (x - 2 log(1 - e^x)))/2 - Li_2(e^x)
x (x - 2 log(1 - e^x)) = x x + x (-2 log(1 - e^x)):
(x x - 2 x log(1 - e^x))/2 - Li_2(e^x)
x x = x^2:
(x^2 - 2 x log(1 - e^x))/2 - Li_2(e^x)
(x^2 - 2 x log(1 - e^x))/2 = x^2/2 - (2 x log(1 - e^x))/2:
(x^2)/2 - (2 x log(1 - e^x))/2 - Li_2(e^x)
(-2)/2 = (2 (-1))/2 = -1:
Answer: |
| x^2/2 + -1 x log(1 - e^x) - Li_2(e^x)
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