Question

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

Answer 1

This is hard...

`int1/sqrt(1+sqrt(1+x^2))dx`

We need to delete all this square !!

For this let's :

`u = sqrt(1+x^2)`

`x = sqrt(u^2-1)`

`du = x/sqrt(1+x^2)dx`

`dx=sqrt(1+x^2)/xdu`

Integral become :

`intsqrt(1+x^2)/(x*sqrt(1+sqrt(1+x^2)))du`

`intu/(sqrt(u^2-1)*sqrt(u+1))du`

Multiply numerator and denominator by`sqrt(u-1)` and don't forget that `(u+1)(u-1) = u^2-1` so we have :

`(usqrt(u-1))/(u^2-1)`

Let's `w = sqrt(u-1)`

`dw = 1/(2sqrt(u-1))du`

`du = 2sqrt(u-1)dw`

`2int(usqrt(u-1))/(u^2-1)sqrt(u-1)dw`

`w^2+1=u`

`(w^2+1)^2=u^2`

`2int((w^2+1)w^2)/((w^2+1)^2-1)dw`

`2int(w^4+w^2)/(w^4+2w^2)dw`

`2int(w^2(w^2+1))/(w^2(w^2+2))dw`

`2int(w^2+1)/(w^2+2)dw`

`2int(w^2+2-1)/(w^2+2)dw`

`2int1dw-2int1/(w^2+2)dw`

`2int1dw-int1/(1/2w^2+1)dw`

Let's `t =1/sqrt(2)w`

`t^2=1/2w^2`

`dt = 1/sqrt(2)dw`

`[2w]- sqrt(2)int1/(t^2+1)dt`

`[2w]-[sqrt(2)arctan(t)]+C`

Substitute back...

`[2w]-[sqrt(2)arctan(1/sqrt(2)w)]+C`

`[2sqrt(u-1)]-[sqrt(2)arctan(1/sqrt(2)sqrt(u-1))]+C`

`[2sqrt(sqrt(x^2+1)-1)]-[sqrt(2)arctan(1/sqrt(2)sqrt(sqrt(x^2+1)-1))]+C`