求积分:∫(x^3+1/x^3-x^2)dx
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求积分:∫(x^3+1/x^3-x^2)dx
![求积分:∫(x^3+1/x^3-x^2)dx](/uploads/image/z/1356790-22-0.jpg?t=%E6%B1%82%E7%A7%AF%E5%88%86%3A%E2%88%AB%28x%5E3%2B1%2Fx%5E3-x%5E2%29dx)
∫(x^3+1)dx/(x^3-x^2) = ∫(x^3-x^2+x^2+1)dx/(x^3-x^2)
= ∫[1+(x^2+1)/(x^3-x^2)]dx
= ∫[1+2/(x-1)-1/x-1/x^2]dx
= x+2ln|x-1|-ln|x|+1/x+C
再问:![](http://img.wesiedu.com/upload/4/a9/4a9c028485e5253ce0047b2a88f0685a.jpg)
再问: 拜托了
再问: 😊
再答: (1) 令 x=sect, 则 I = ∫(tant)^2dt
= [tant-t] = 2-arctan2
(2) I = (1/9)∫[√(x+9)+√x]dx
= (2/27)[(x+9)^(3/2)-x^(3/2)] = 122/27
= ∫[1+(x^2+1)/(x^3-x^2)]dx
= ∫[1+2/(x-1)-1/x-1/x^2]dx
= x+2ln|x-1|-ln|x|+1/x+C
再问:
![](http://img.wesiedu.com/upload/4/a9/4a9c028485e5253ce0047b2a88f0685a.jpg)
再问: 拜托了
再问: 😊
再答: (1) 令 x=sect, 则 I = ∫(tant)^2dt
= [tant-t] = 2-arctan2
(2) I = (1/9)∫[√(x+9)+√x]dx
= (2/27)[(x+9)^(3/2)-x^(3/2)] = 122/27