∫1/x(x^2+1)^(1/2)dx=?请朋友们说下过程吧,

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∫1/x(x^2+1)^(1/2)dx=?请朋友们说下过程吧,
∫1/x(x^2+1)^(1/2)dx=?请朋友们说下过程吧,

∫1/x(x^2+1)^(1/2)dx=?请朋友们说下过程吧,
∫ dx/[x√(x^2+1)]
let
x= tany
dx = (secy)^2dy
∫ (secy/tany) dy
=∫ cscy dy
=ln|cscy-coty| + C
=ln|√(x^2+1)/x - 1/x | + C

答：
设t=√(x^2+1)，x=√(t^2-1)
∫ {1/[x√(x^2+1)]} dx
=∫ {1/[t*√(t^2-1)] d[√(t^2-1)]
=∫ {1/[t*√(t^2-1)]}*(1/2)*[2t/√(t^2-1)] dt
=∫ [1/(t^2-1)] dt
=(1/2) ∫ [1/(t-1)-1/(t+1)] dt
=(1/2)*[ln|t-1|-ln|t+1|+C
=(1/2)*ln|(t-1)/(t+1)|+C
=(1/2)*ln|[√(x^2+1)-1]/[√(x^2+1)+1]|+C

∫ x/(1+X^2)dx= ∫(x+1/x)^2dx=? ∫(1+x)/(X^2)dx=∫ [(1+x)/(X^2)]dx得什么? ∫[dx/(e^x(1+e^2x)]dx ∫1+2x/x(1+x)*dx∫1+2x/x(1+x) * dx 1/(x+x^2)dx ∫(x-1)^2dx, ∫x^1/2dx ∫x[x/[(2a-x)]^(1/2)dx=? ∫1/x^2+x+1dx ∫1/(x^2+x+1)dx ∫dx/x^2(1-x^2) ∫dx/x^2(1+x^2) ∫X^2/1-x^2 dx. ∫X^2/1-x^2 dx. ∫ (x+arctanx)/(1+x^2) dx ∫（x^2+1/x^4)dx ∫2 -1|x²-x|dx