Integration Problem

(a).Put u=1/(x+1), du=(-1)(x+1)^(-2) =-1/(x+1)²dxPut dv=(e^x)dx, v=e^xSo, sub. u, v, du, dv into ∫udv=uv-∫vdu ∫(e^x)/(x+1)dx=[1/(x+1)](e^x)-∫(e^x)[-1/(x+1)²dx]=(e^x)/(x+1)+∫(e^x)/(x+1)²dx (b).∫x(e^x)/(x+1)²dx=∫(x+1-1)(e^x)/(x+1)²dx=∫[(x+1)(e^x)/(x+1)²-(e^x)/(x+1)²]dx=∫[(e^x)/(x+1)-(e^x)/(x+1)²]dx=∫(e^x)/(x+1)dx-∫(e^x)/(x+1)²dx=(e^x)/(x+1)+∫(e^x)/(x+1)²dx-∫(e^x)/(x+1)²dx=(e^x)/(x+1)