∫1/(sec^2x+1)dx等于多少?
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∫1/(sec^2x+1)dx等于多少?∫1/(sec^2x+1)dx等于多少?∫1/(sec^2x+1)dx等于多少?∫1/(sec²x+1)dx=∫[(1+sec²x)-sec
∫1/(sec^2x+1)dx等于多少?
∫1/(sec^2x+1)dx等于多少?
∫1/(sec^2x+1)dx等于多少?
∫ 1/(sec²x + 1) dx
= ∫ [(1 + sec²x) - sec²x]/(sec²x + 1) dx
= ∫ dx - ∫ sec²x/(sec²x + 1) dx
= x - ∫ d(tanx)/(tan²x + 2),恒等式1 + tan²x = sec²x
= x - (1/√2)arctan[(tanx)/√2] + C
原式=∫ (2x+1)/√[(x+1)²+1] dx
令x+1=tanu,则√[(x+1)²+1]=secu,dx=sec²udu
=∫ [(2tanu-1)/secu]sec²u du
=∫ [2tanusecu-secu) du
=2secu - ln|secu+tanu| + C
=2√(x²+2x+2) - ln|√(x²+2x+2) + x + 1| + C
选为满意答案
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