Question

一个高数不定积分的题，教我怎么做

∫｛x^2e^x/(x+2)^2}dx=???

这是高等数学的题，求详细解答，分母是(x+2)^2

Answer 1

∫(x^2e^x/(2+x)^2 dx

∫[x^2*e^x/(x+2)^2]dx 换元积分
=-∫(x^2*e^x)d[1/(x+2)] 分部积分
=-{x^2*e^x/(x+2)-∫[1/(x+2)]d(x^2*e^x)}
=[-x^2*e^x/(x+2)]+∫[1/(x+2)]*(2x*e^x+x^2*e^x)dx
=[-x^2*e^x/(x+2)]+∫[1/(x+2)]*x*e^x*(x+2)dx
=[-x^2*e^x/(x+2)]+∫x*e^xdx
=[-x^2*e^x/(x+2)]+∫xd(e^x) 再分部积分
=[-x^2*e^x/(x+2)]+[x*e^x-∫e^xdx]
=[-x^2*e^x/(x+2)]+x*e^x-e^x+C 合并同类项
=[-x^2/(x+2)+(x-1)]*e^x+C
=[(x-2)/(x+2)]*e^x+C

Answer 2

∫x^2e^x/(2+x)^2 dx

∫[x^2*e^x/(x+2)^2]dx
=-∫(x^2*e^x)d[1/(x+2)]
=-{x^2*e^x/(x+2)-∫[1/(x+2)]d(x^2*e^x)}
=[-x^2*e^x/(x+2)]+∫[1/(x+2)]*(2x*e^x+x^2*e^x)dx
=[-x^2*e^x/(x+2)]+∫[1/(x+2)]*x*e^x*(x+2)dx
=[-x^2*e^x/(x+2)]+∫x*e^xdx
=[-x^2*e^x/(x+2)]+∫xd(e^x)
=[-x^2*e^x/(x+2)]+[x*e^x-∫e^xdx]
=[-x^2*e^x/(x+2)]+x*e^x-e^x+C
=[-x^2/(x+2)+(x-1)]*e^x+C
=[(x-2)/(x+2)]*e^x+C