∫[(3x+1)/(4+x^2)^(1/2)]dx高等数学,

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∫[(3x+1)/(4+x^2)^(1/2)]dx高等数学,
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∫[(3x+1)/(4+x^2)^(1/2)]dx高等数学,
∫[(3x+1)/(4+x^2)^(1/2)]dx高等数学,

∫[(3x+1)/(4+x^2)^(1/2)]dx高等数学,
x=2 tant,dx = 2(sect)^2 dt ,(4+x^2)^(1/2) = 2 sect
化为三角函数有理式的积分
∫ ( 6 tant +1) sect dt = ∫ 6 sect tant dt + ∫ sect dt
= 6sect + ln| sect +tant | + C
= .
积分的题目也是有规律的,根据被积函数的特点,选择如何去掉根号,化为有理函数.

先把X=2tanx 换得算，得到3sinx+1/2secx3，3sinx直接求积分。。。1/2secx 在再用分部积分算

∫[(3x+1)/(4+x^2)^(1/2)]dx=∫3x/(4+x^2)^(1/2)dx+∫1/(4+x^2)^(1/2)dx
=3(4+x^2)^(1/2)+ ln(x+(x^2+4)^2)

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