这道积分怎么算？
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∫(0->+∞) 3x/(x+1)^4 dx
=3∫(0->+∞) dx/(x+1)^3 - 3∫(0->+∞) dx/(x+1)^4
=3[ -1/(2(x+1)^2) + 1/(3(x+1)^3) ](0->+∞)
=3(1/2 -1/3)
=1/2
3x^2
=3(x+1)^2 - 6x-3
=3(x+1)^2 - 6(x+1) +3
∫(0->+∞) 3x^2/(x+1)^4 dx
=3∫(0->+∞) dx/(x+1)^2 -6∫(0->+∞) dx/(x+1)^3 + 3∫(0->+∞) dx/(x+1)^4
=3[ -1/(x+1) +1/(x+1)^2 - 1/(3(x+1)^3) ] |(0->+∞)
=3( 1 - 1 +1/3)
=1
=3∫(0->+∞) dx/(x+1)^3 - 3∫(0->+∞) dx/(x+1)^4
=3[ -1/(2(x+1)^2) + 1/(3(x+1)^3) ](0->+∞)
=3(1/2 -1/3)
=1/2
3x^2
=3(x+1)^2 - 6x-3
=3(x+1)^2 - 6(x+1) +3
∫(0->+∞) 3x^2/(x+1)^4 dx
=3∫(0->+∞) dx/(x+1)^2 -6∫(0->+∞) dx/(x+1)^3 + 3∫(0->+∞) dx/(x+1)^4
=3[ -1/(x+1) +1/(x+1)^2 - 1/(3(x+1)^3) ] |(0->+∞)
=3( 1 - 1 +1/3)
=1
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