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

Jul 1, 2015

The answer is $\left[\frac{1}{2} \ln \left(\frac{1}{\sqrt{{x}^{2} + 1}} - 1\right) + \sqrt{{x}^{2} + 1} - \frac{1}{2} \ln \left(\frac{1}{\sqrt{{x}^{2} + 1}} + 1\right)\right] + C$

#### Explanation:

let's $x = \tan \left(u\right)$

$\mathrm{dx} = \frac{1}{\cos} ^ 2 \left(u\right) \mathrm{du}$

the integral become :

$\int \frac{\frac{1}{\cos} \left(u\right)}{\tan} \left(u\right) \cdot \frac{1}{\cos} ^ 2 \left(u\right) \mathrm{du}$

$\int \frac{1}{\tan} \left(u\right) \cdot \frac{1}{\cos} ^ 3 \left(u\right) \mathrm{du}$

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

$\int \sin \frac{u}{\sin} {\left(u\right)}^{2} {\cos}^{2} \left(u\right)$

$t = \cos \left(u\right)$

$\mathrm{dt} = - \sin \left(u\right)$

$- \int \frac{1}{\left(1 - {t}^{2}\right) {t}^{2}}$

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

with partial fraction we get

$- \int - \frac{1}{2 \left(t - 1\right)} + \frac{1}{t} ^ 2 + \frac{1}{2 \left(1 + t\right)}$

$\left[\frac{1}{2} \ln \left(t - 1\right) + \frac{1}{t} - \frac{1}{2} \ln \left(t + 1\right)\right] + C$

since $t = \cos \left(u\right)$

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

remember $x = \tan \left(u\right)$

so $\arctan \left(x\right) = u$

$\cos \left(\arctan \left(x\right)\right) = \cos \left(u\right)$

but $\cos \left(\arctan \left(x\right)\right) = \frac{1}{\sqrt{{x}^{2} + 1}}$

FINALLY :

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