Evaluate following integral: ∫ [1 / {x √(a² + x²)}] dx?

√a^2+x^2=t so x^2=a^2-t^2 and x=√a^2-t^2

1/2(√a^2+x^2 )(2x)dx=dt, x/(√a^2+x^2 )dx=dt

i=Int xdx/x^2(√a^2+x^2)= Int dt/(√a^2-t^2)=1/2a log | (t+a)/(t-a)| +c put value of t and get result

x = a * tan(t)

dx = a * sec(t)^2 * dt

dx / (x * sqrt(a^2 + x^2)) =>

a * sec(t)^2 * dt / (a * tan(t) * sqrt(a^2 + a^2 * tan(t)^2)) =>

a * sec(t)^2 * dt / (a * tan(t) * sqrt(a^2 * sec(t)^2)) =>

a * sec(t)^2 * dt / (a * tan(t) * a * sec(t)) =>

sec(t) * dt / (a * tan(t)) =>

(1/a) * dt / sin(t) =>

(1/a) * csc(t) * dt

Integrate

(-1/a) * ln|csc(t) + cot(t)| + C

x = a * tan(t)

x/a = tan(t)

a/x = cot(t)

csc(t)^2 - cot(t)^2 = 1

csc(t)^2 - (a/x)^2 = 1

csc(t)^2 = (a^2 + x^2) / x^2

csc(t) = sqrt(a^2 + x^2) / x

(-1/a) * ln|csc(t) + cot(t)| + C =>

(-1/a) * ln|(1/x) * (a + sqrt(a^2 + x^2))| + C =>

(1/a) * (ln|x| - ln|a + sqrt(a^2 + x^2)|) + C

It appears to be continuous, see graph

https://www.youtube.com/watch?v=_4AghQh9NaE

answer is in the image