作业帮求积分∫(xe^x)/{[(e^2-1)]^(1/2)}dx
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![求积分∫(xe^x)/{[(e^2-1)]^(1/2)}dx](/uploads/image/z/2650998-30-8.jpg?t=%E6%B1%82%E7%A7%AF%E5%88%86%E2%88%AB%28xe%5Ex%29%2F%7B%5B%28e%5E2-1%29%5D%5E%281%2F2%29%7Ddx)
求积分∫(xe^x)/{[(e^2-1)]^(1/2)}dx
求积分∫(xe^x)/{[(e^2-1)]^(1/2)}dx
求积分∫(xe^x)/{[(e^2-1)]^(1/2)}dx
integral (e^x x)/sqrt(e^x-1) dx
For the integrand (e^x x)/sqrt(e^x-1), substitute u = sqrt(e^x-1) and du = e^x/(2 sqrt(e^x-1)) dx:
= integral 2 log(u^2+1) du
Factor out constants:
= 2 integral log(u^2+1) du
For the integrand log(u^2+1), integrate by parts, integral f dg = f g- integral g df, where
f = log(u^2+1), dg = du,
df = (2 u)/(u^2+1) du, g = u:
= 2 u log(u^2+1)-2 integral (2 u^2)/(u^2+1) du
Factor out constants:
= 2 u log(u^2+1)-4 integral u^2/(u^2+1) du
For the integrand u^2/(u^2+1), do long division:
= 2 u log(u^2+1)-4 integral (1-1/(u^2+1)) du
Integrate the sum term by term and factor out constants:
= 2 u log(u^2+1)+4 integral 1/(u^2+1) du-4 integral 1 du
The integral of 1/(u^2+1) is tan^(-1)(u):
= 2 u log(u^2+1)+4 tan^(-1)(u)-4 integral 1 du
The integral of 1 is u:
= 2 u log(u^2+1)-4 u+4 tan^(-1)(u)+constant
Substitute back for u = sqrt(e^x-1):
= -4 sqrt(e^x-1)+2 sqrt(e^x-1) log(e^x)+4 tan^(-1)(sqrt(e^x-1))+constant
Factor the answer a different way:
Answer: |
| = 2 sqrt(e^x-1) (log(e^x)-2)+4 tan^(-1)(sqrt(e^x-1))+constant
希望采纳.