Question

根号下（x^2＋4）/x^2 dx的不定积分 求详细解答过程

Answer 1

方法如下，请作参考：

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Answer 2

设 x = 2tant， 则 dx = 2(sect)^2dt
I = ∫[√(x^2+4)/x^2]dx = ∫4(sect)^3dt/[4(tant)^2]
= ∫dt/[cost(sint)^2] = ∫dsint/[(cost)^2(sint)^2], 记 u = sint
I = ∫du/[u^2(1-u^2)] = ∫[1/u^2 + (1/2)[1/(1-u)+1/(1+u)]du
= -1/u + (1/2)ln|(1+u)/(1-u)| + C
= -1/sint + (1/2)ln|(1+sint)/(1-sint)| + C
= -1/sint + (1/2)ln|(1+sint)/(1-sint)| + C
= -√(x^2+4)/x + (1/2)ln|[1+x/√(x^2+4)]/[1-x/√(x^2+4)]| + C
= -√(x^2+4)/x + (1/2)ln|[√(x^2+4)+x]/[√(x^2+4)-x]| + C
= -√(x^2+4)/x + ln|[√(x^2+4)+x]/2| + C
= -√(x^2+4)/x + ln[√(x^2+4)+x] + C1