Question

若f(x)的一个原函数为(x-1)e^x ,求 ∫xf ' (x)dx ∫（1/x)f ( lnx )dx？

Answer 1

∫ f(x) dx = (x - 1)e^x = xe^x - e^x
f(x) = (xe^x - e^x)' = (xe^x + e^x) - e^x = xe^x
∫ xf'(x) dx
= ∫ x d[f(x)]
= xf(x) - ∫ f(x) dx
= x(xe^x) - (xe^x - e^x) + C
= x²e^x - xe^x + e^x + C
= (x² - x +1)e^x + C
∫ (1/x)f(lnx) dx
= ∫ f(lnx) d(lnx) = ∫ f(u) du,u = lnx
= (u - 1)e^u + C
= (lnx - 1)e^(lnx) + C
= x(lnx - 1) + C,6,0.0,1,f ' (x)=e^x +(x-1)e^x=x*e^x
∫xf ' (x)dx ∫（1/x)f ( lnx )dx=∫e^xdx ∫x（Inx-1)dx=∫x（1-e^x)dx=1/2x^2-xe^x+e^x+c 其中 c为常数,0,