Integral of 1/sqrt(1+x^2)

Integral of 1/sqrt(1+x^2)

Indefinite integral
integral 1/sqrt(1 + x^2) dx = sinh^(-1)(x) + constant
Alternate form of the integral
log(sqrt(x^2 + 1) + x) + constant
Series expansion of the integral at x=0
x – x^3/6 + (3 x^5)/40 + O(x^6)
(Taylor series)
Series expansion of the integral at x=-i
-i (pi/2 + (-1)^(floor(((3 pi)/2 – arg(x + i))/(2 pi))) ((-1)^(3/4) sqrt(2) sqrt(x + i) + ((-1)^(1/4) (x + i)^(3/2))/(6 sqrt(2)) – (3 (-1)^(3/4) (x + i)^(5/2))/(80 sqrt(2)) – (5 (-1)^(1/4) (x + i)^(7/2))/(448 sqrt(2)) + (35 (-1)^(3/4) (x + i)^(9/2))/(9216 sqrt(2)) + O((x + i)^(11/2))))
Series expansion of the integral at x=i
1/2 i (pi + (-1)^(floor((pi – 2 arg(x – i))/(4 pi))) (-(2 + 2 i) sqrt(x – i) + (1/6 – i/6) (x – i)^(3/2) + (3/80 + (3 i)/80) (x – i)^(5/2) – (5/448 – (5 i)/448) (x – i)^(7/2) – (35/9216 + (35 i)/9216) (x – i)^(9/2) + O((x – i)^(11/2))))
Series expansion of the integral at x=∞
log(2 x) + 1/(4 x^2) – 3/(32 x^4) + O((1/x)^6)
(generalized Puiseux series)

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