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

Mar 20, 2017

$\int \sqrt{1 + \frac{1}{x} ^ 2} \mathrm{dx} = \sqrt{{x}^{2} + 1} + \frac{1}{2} \ln \left\mid \frac{\sqrt{{x}^{2} + 1} - 1}{\sqrt{{x}^{2} + 1} + 1} \right\mid + C$

#### Explanation:

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

Letting $u = \sqrt{{x}^{2} + 1}$ reveals that $\mathrm{du} = \frac{x}{\sqrt{{x}^{2} + 1}} \mathrm{dx}$. Then the integral can be manipulated to become:

$\int \frac{\sqrt{{x}^{2} + 1}}{x} \mathrm{dx} = \int \frac{x \left({x}^{2} + 1\right)}{{x}^{2} \sqrt{{x}^{2} + 1}} \mathrm{dx} = \int \frac{{x}^{2} + 1}{x} ^ 2 \left(\frac{x}{\sqrt{{x}^{2} + 1}} \mathrm{dx}\right)$

Note that ${u}^{2} = {x}^{2} + 1$ and ${u}^{2} - 1 = {x}^{2}$:

$\int \frac{{x}^{2} + 1}{x} ^ 2 \left(\frac{x}{\sqrt{{x}^{2} + 1}} \mathrm{dx}\right) = \int {u}^{2} / \left({u}^{2} - 1\right) \mathrm{du}$

Rewriting the integrand as $\frac{{u}^{2} - 1 + 1}{{u}^{2} - 1} = 1 + \frac{1}{{u}^{2} - 1}$:

$\int {u}^{2} / \left({u}^{2} - 1\right) \mathrm{du} = \int \mathrm{du} + \int \frac{1}{{u}^{2} - 1} \mathrm{du} = u + \int \frac{1}{{u}^{2} - 1} \mathrm{du}$

You can perform partial fraction decomposition on $\frac{1}{{u}^{2} - 1}$ to see that $\frac{1}{{u}^{2} - 1} = \frac{1}{2 \left(u - 1\right)} - \frac{1}{2 \left(u + 1\right)}$:

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

Both of which you could perform a substitution on, or just recognize them for natural log integrals:

$u + \frac{1}{2} \int \frac{1}{u - 1} \mathrm{du} - \frac{1}{2} \int \frac{1}{u + 1} \mathrm{du} = u + \frac{1}{2} \ln \left\mid u - 1 \right\mid - \frac{1}{2} \ln \left\mid u + 1 \right\mid$

Combining:

$= u + \frac{1}{2} \ln \left\mid u - 1 \right\mid - \frac{1}{2} \ln \left\mid u + 1 \right\mid = u + \frac{1}{2} \ln \left\mid \frac{u - 1}{u + 1} \right\mid$

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

$u + \frac{1}{2} \ln \left\mid \frac{u - 1}{u + 1} \right\mid = \sqrt{{x}^{2} + 1} + \frac{1}{2} \ln \left\mid \frac{\sqrt{{x}^{2} + 1} - 1}{\sqrt{{x}^{2} + 1} + 1} \right\mid$

Or:

$\int \sqrt{1 + \frac{1}{x} ^ 2} \mathrm{dx} = \sqrt{{x}^{2} + 1} + \frac{1}{2} \ln \left\mid \frac{\sqrt{{x}^{2} + 1} - 1}{\sqrt{{x}^{2} + 1} + 1} \right\mid + C$

Jun 22, 2017

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

Let $x = \cot \theta$. This implies that $\mathrm{dx} = - {\csc}^{2} \theta d \theta$ and that $1 + \frac{1}{x} ^ 2 = 1 + {\tan}^{2} \theta = {\sec}^{2} \theta$. Then:

$I = \int \sqrt{{\sec}^{2} \theta} \left(- {\csc}^{2} \theta d \theta\right) = - \int \sec \theta \left(1 + {\cot}^{2} \theta\right) d \theta$

$\textcolor{w h i t e}{I} = - \int \left(\sec \theta + \cot \theta \csc \theta\right) d \theta = - \ln \left\mid \sec \theta + \tan \theta \right\mid + \csc \theta$

Rewriting in terms of $\cot \theta$:

$I = - \ln \left\mid \sqrt{1 + \frac{1}{\cot} ^ 2 \theta} + \frac{1}{\cot} \theta \right\mid + \sqrt{1 + {\cot}^{2} \theta}$

$\textcolor{w h i t e}{I} = - \ln \left\mid \sqrt{1 + \frac{1}{x} ^ 2} + \frac{1}{x} \right\mid + \sqrt{1 + {x}^{2}}$

$\textcolor{w h i t e}{I} = \ln \left\mid \frac{x}{1 + \sqrt{1 + {x}^{2}}} \right\mid + \sqrt{1 + {x}^{2}} + C$