作业帮高数简单微分计算∫[（2x+1)/(x^2+2x+2]）dx
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高数简单微分计算∫[（2x+1)/(x^2+2x+2]）dx
高数简单微分计算∫[（2x+1)/(x^2+2x+2]）dx
高数简单微分计算∫[（2x+1)/(x^2+2x+2]）dx
∫(2x+1)dx/(x^2+2x+2)
=∫d(x^2+2x+2)/(x^2+2x+2)-∫dx/(x^2+2x+2)
=ln(x^2+2x+2)-arctan(x+1)+C
(2x + 1) / (x^2 + 2x + 2) = (2(x + 1) - 1) / ((x + 1)^2 + 1) = 2(x+1) / ((x + 1)^2 + 1) - 1 / ((x + 1)^2 + 1)
∫ [(2x+1)/(x^2+2x+2])dx = 2∫[ (x+1) / ((x + 1)^2 + 1)] dx- ∫ [1 / ((x + 1)^2 + 1)] ...
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(2x + 1) / (x^2 + 2x + 2) = (2(x + 1) - 1) / ((x + 1)^2 + 1) = 2(x+1) / ((x + 1)^2 + 1) - 1 / ((x + 1)^2 + 1)
∫ [(2x+1)/(x^2+2x+2])dx = 2∫[ (x+1) / ((x + 1)^2 + 1)] dx- ∫ [1 / ((x + 1)^2 + 1)] dx
= ∫ [ 1 / ((x + 1)^2 + 1)] d[(x+1)^2 + 1] - ∫ [1 / ((x + 1)^2 + 1)] d(x+1)
= 2ln((x + 1)^2 + 1) - arctan(x+1) + C
= 2ln(x^2 + 2x + 2) - arctan(x+1) + C
C为任意常数.
用到了， ∫ 1 / (1 + t^2) dt = arctan t + C.
望采纳~~
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