一个高数不定积分的题,∫{x^2e^x/(x+2)^2}dx=?这是高等数学的题,分母是(x+2)^2

邓荣君 1年前 已收到2个回答 举报

礼书 花朵

共回答了19个问题采纳率:78.9% 举报

∫(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

1年前

5

ffandl 幼苗

共回答了16个问题 举报

∫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+...

1年前

1
可能相似的问题
Copyright © 2024 YULUCN.COM - 雨露学习互助 - 17 q. 1.574 s. - webmaster@yulucn.com