高等数学。 请问图中题怎么做??
1个回答
展开全部
令 x = tanu, 则
I = ∫(secu)^2du/[(tanu)^4secu]
= ∫secudu/(tanu)^4 = ∫(cosu)^3du/(sinu)^4
= ∫[1-(sinu)^2]dsinu/(sinu)^4
= ∫[1/(sinu)^4 - 1/(sinu)^2]dsinu
= -(1/3)/(sinu)^3 + 1/sinu + C
= √(1+x^2)/x - (1/3)(1+x^2)^(3/2)/x^3 + C
I = ∫(secu)^2du/[(tanu)^4secu]
= ∫secudu/(tanu)^4 = ∫(cosu)^3du/(sinu)^4
= ∫[1-(sinu)^2]dsinu/(sinu)^4
= ∫[1/(sinu)^4 - 1/(sinu)^2]dsinu
= -(1/3)/(sinu)^3 + 1/sinu + C
= √(1+x^2)/x - (1/3)(1+x^2)^(3/2)/x^3 + C
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询