∫2x^2/(x^2+1) 怎么等于2x +∫2/(x^2+1)dx 后面那个2x +是怎么来的

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∫2x^2/(x^2+1)怎么等于2x+∫2/(x^2+1)dx后面那个2x+是怎么来的∫2x^2/(x^2+1)怎么等于2x+∫2/(x^2+1)dx后面那个2x+是怎么来的∫2x^2/(x^2+1

∫2x^2/(x^2+1) 怎么等于2x +∫2/(x^2+1)dx 后面那个2x +是怎么来的
∫2x^2/(x^2+1) 怎么等于2x +∫2/(x^2+1)dx 后面那个2x +是怎么来的

∫2x^2/(x^2+1) 怎么等于2x +∫2/(x^2+1)dx 后面那个2x +是怎么来的
2x^2/(x^2+1)
=(2x^2+2-2)/(x^2+1)
=[2(x^2+1)-2]/(x^2+1)
=2(x^2+1)/(x^2+1)-2/(x^2+1)
=2-2/(x^2+1)
所以原式=∫2dx-∫2dx/(x^2+1)
=2x-∫2dx/(x^2+1)
=2x-2arctanx+C