Question

不定积分∫(x²+1)deˣ=

Answer 1

∫(x^2+1) de^x

=(x^2+1).e^x - 源友∫2xe^x dx

=(x^2+1).e^x - ∫毕顷2x de^x

=(x^2+1).e^x - 2xe^x +2∫雹数槐e^x dx

=(x^2+1).e^x - 2xe^x +2e^x +C

Answer 2

需要多纤槐启次毁如使用明卖分部积分来求！
∫(x^2+1)de^x
=(x^2+1)e^x-∫e^xd(x^2+1)
=(x^2+1)e^x-2∫xe^xdx
=(x^2+1)e^x-2xe^x+e^x+C