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let
u=(x+1)^(1/漏锋嫌6)
du =(1/6)(x+1)^(-5/基闭6) dx
dx =6u^5 du
u^3
=u^2.(u+1) -u^2
=u^2.(u+1) -u(u+1) +u
=u^2.(u+1) -u(u+1) +(1+u) -1
∫ dx/[ √(x+1) +(x+1)^(1/3) ]
=∫ 6u^5 du/( u^3 +u^2 )
=6∫ u^3/( u +1 ) du
=6∫返手 [ u^2-u +1 - 1/( u +1 )] du
=6[ (1/3)u^3 -(1/2)u^2 +u - ln|1+u| ] +C
=6[ (1/3)(x+1)^(1/2) -(1/2)(x+1)^(1/3) +(x+1)^(1/6) - ln|1+(x+1)^(1/6)| ] +C
u=(x+1)^(1/漏锋嫌6)
du =(1/6)(x+1)^(-5/基闭6) dx
dx =6u^5 du
u^3
=u^2.(u+1) -u^2
=u^2.(u+1) -u(u+1) +u
=u^2.(u+1) -u(u+1) +(1+u) -1
∫ dx/[ √(x+1) +(x+1)^(1/3) ]
=∫ 6u^5 du/( u^3 +u^2 )
=6∫ u^3/( u +1 ) du
=6∫返手 [ u^2-u +1 - 1/( u +1 )] du
=6[ (1/3)u^3 -(1/2)u^2 +u - ln|1+u| ] +C
=6[ (1/3)(x+1)^(1/2) -(1/2)(x+1)^(1/3) +(x+1)^(1/6) - ln|1+(x+1)^(1/6)| ] +C
追问
你的答案跟我一样,可是后面答案不是这个
追答
答案有点奇怪,把答案求导看看能不能变回 1/[ √(x+1) +(x+1)^(1/3) ], 这个我不能肯定?
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