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

Question

Isabella Ackerman · Accepted Answer

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#

Cameron Alexander · Answer

undefined To find the integral of $ \frac{1}{\sqrt{1+\sqrt{1+x^2}}} $, we can use the substitution method. Let $ u = \sqrt{1+x^2} $, then $ du = \frac{x}{\sqrt{1+x^2}} \, dx $.

Substituting $ u = \sqrt{1+x^2} $ into the integral, we have:

$$ \int \frac{1}{\sqrt{1+\sqrt{1+x^2}}} \, dx = \int \frac{1}{\sqrt{u}} \cdot \frac{1}{u} \, du $$

This simplifies to:

$$ \int \frac{1}{u^{3/2}} \, du $$

Now, integrating $ \frac{1}{u^{3/2}} $, we get:

$$ \int \frac{1}{u^{3/2}} \, du = \int u^{-3/2} \, du = \frac{u^{-1/2}}{-1/2} + C $$

Where $ C $ is the constant of integration.

Substituting back for $ u = \sqrt{1+x^2} $, we get:

$$ -2\sqrt{1+x^2} + C $$

Therefore, the integral of $ \frac{1}{\sqrt{1+\sqrt{1+x^2}}} $ is $ -2\sqrt{1+x^2} + C $, where $ C $ is the constant of integration.