作者m870171 (super)
看板trans_math
标题Re: [积分] 请问各位大大...
时间Fri Mar 3 16:26:36 2006
※ 引述《hairgirl (你好就好.............)》之铭言：
: ∫√x(x+1)^2dx
: 请各位大大帮忙一下罗，谢谢...
第一种 ∫(√x)(x+1)^2dx
令x=t^2
= ∫t(t^2+1)^2d(t^2)
又 dt^2=2t dt
= ∫2t^2(t^4 + 2t^2 +1)dt
= 2∫(t^6 + 2t^4 + t^2)dt
= 2(1/7t^7 + 2/5t^5 + 1/3t^3 + c)
= 2/7t^7 + 4/5t^5 + 2/3t^3 + C
第二种 ∫√{x(x+1)^2}dx
令x=t^2
= ∫t(t^2 + 1)d(t^2)
又 dt^2=2t dt
= ∫2t^2(t^2 + 1)d(t)
= 2∫(t^4 +t^2) dt
= 2(1/5t^5 + 1/3t^3 + c)
= 2/5t^5 + 2/3t^3 + C
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