How do you find the integral of #(sqrt(1+x^2)/x)#?

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Answer 1 · 2015-07-01T17:54:21

The answer is #[1/2ln(1/sqrt(x^2+1)-1)+sqrt(x^2+1)-1/2ln(1/sqrt(x^2+1)+1)]+C#

let's #x = tan(u)#

#dx = 1/cos^2(u)du#

the integral become :

#int (1/cos(u))/tan(u)*1/cos^2(u) du#

#int 1/tan(u)*1/cos^3(u) du#

#int 1/(sin(u)cos^2(u)#

#int sin(u)/sin(u)^2cos^2(u)#

#t = cos(u)#

#dt = -sin(u)#

#-int1/((1-t^2)t^2)#

#-int1/((1+t)(1-t)t^2#

with partial fraction we get

#-int-1/(2 (t-1)) + 1/t^2 + 1/(2 (1 + t))#

#[1/2ln(t-1)+1/t-1/2ln(t+1)]+C#

since #t = cos(u)#

#[1/2ln(cos(u)-1)+1/cos(u)-1/2ln(cos(u)+1)]+C#

remember #x = tan(u)#

so #arctan(x) = u#

#cos(arctan(x)) = cos(u)#

but #cos(arctan(x)) = 1/sqrt(x^2+1)#

FINALLY :

#[1/2ln(1/sqrt(x^2+1)-1)+sqrt(x^2+1)-1/2ln(1/sqrt(x^2+1)+1)]+C#