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

May 13, 2015

This is hard...

$\int \frac{1}{\sqrt{1 + \sqrt{1 + {x}^{2}}}} \mathrm{dx}$

We need to delete all this square !!

For this let's :

$u = \sqrt{1 + {x}^{2}}$

$x = \sqrt{{u}^{2} - 1}$

$\mathrm{du} = \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx}$

$\mathrm{dx} = \frac{\sqrt{1 + {x}^{2}}}{x} \mathrm{du}$

Integral become :

$\int \frac{\sqrt{1 + {x}^{2}}}{x \cdot \sqrt{1 + \sqrt{1 + {x}^{2}}}} \mathrm{du}$

$\int \frac{u}{\sqrt{{u}^{2} - 1} \cdot \sqrt{u + 1}} \mathrm{du}$

Multiply numerator and denominator by$\sqrt{u - 1}$ and don't forget that $\left(u + 1\right) \left(u - 1\right) = {u}^{2} - 1$ so we have :

$\frac{u \sqrt{u - 1}}{{u}^{2} - 1}$

Let's $w = \sqrt{u - 1}$

$\mathrm{dw} = \frac{1}{2 \sqrt{u - 1}} \mathrm{du}$

$\mathrm{du} = 2 \sqrt{u - 1} \mathrm{dw}$

$2 \int \frac{u \sqrt{u - 1}}{{u}^{2} - 1} \sqrt{u - 1} \mathrm{dw}$

${w}^{2} + 1 = u$

${\left({w}^{2} + 1\right)}^{2} = {u}^{2}$

$2 \int \frac{\left({w}^{2} + 1\right) {w}^{2}}{{\left({w}^{2} + 1\right)}^{2} - 1} \mathrm{dw}$

$2 \int \frac{{w}^{4} + {w}^{2}}{{w}^{4} + 2 {w}^{2}} \mathrm{dw}$

$2 \int \frac{{w}^{2} \left({w}^{2} + 1\right)}{{w}^{2} \left({w}^{2} + 2\right)} \mathrm{dw}$

$2 \int \frac{{w}^{2} + 1}{{w}^{2} + 2} \mathrm{dw}$

$2 \int \frac{{w}^{2} + 2 - 1}{{w}^{2} + 2} \mathrm{dw}$

$2 \int 1 \mathrm{dw} - 2 \int \frac{1}{{w}^{2} + 2} \mathrm{dw}$

$2 \int 1 \mathrm{dw} - \int \frac{1}{\frac{1}{2} {w}^{2} + 1} \mathrm{dw}$

Let's $t = \frac{1}{\sqrt{2}} w$

${t}^{2} = \frac{1}{2} {w}^{2}$

$\mathrm{dt} = \frac{1}{\sqrt{2}} \mathrm{dw}$

$\left[2 w\right] - \sqrt{2} \int \frac{1}{{t}^{2} + 1} \mathrm{dt}$

$\left[2 w\right] - \left[\sqrt{2} \arctan \left(t\right)\right] + C$

Substitute back...

$\left[2 w\right] - \left[\sqrt{2} \arctan \left(\frac{1}{\sqrt{2}} w\right)\right] + C$

$\left[2 \sqrt{u - 1}\right] - \left[\sqrt{2} \arctan \left(\frac{1}{\sqrt{2}} \sqrt{u - 1}\right)\right] + C$

$\left[2 \sqrt{\sqrt{{x}^{2} + 1} - 1}\right] - \left[\sqrt{2} \arctan \left(\frac{1}{\sqrt{2}} \sqrt{\sqrt{{x}^{2} + 1} - 1}\right)\right] + C$