# How do you evaluate the integral int dx/(x^4+1)?

Aug 16, 2017

$\frac{1}{2 \sqrt{2}} a r c \tan \left(\frac{{x}^{2} - 1}{\sqrt{2} \cdot x}\right) - \frac{1}{4 \sqrt{2}} \ln | \frac{{x}^{2} - \sqrt{2} \cdot x + 1}{{x}^{2} - \sqrt{2} \cdot x + 1} | + C .$

#### Explanation:

First, let us consider the following Integrals :

${I}_{1} = \int \frac{{x}^{2} + 1}{{x}^{4} + 1} \mathrm{dx} , \mathmr{and} , {I}_{2} = \int \frac{{x}^{2} - 1}{{x}^{4} + 1} \mathrm{dx} .$

Now, ${I}_{1} = \int \frac{{x}^{2} \left(1 + \frac{1}{x} ^ 2\right)}{{x}^{2} \left({x}^{2} + \frac{1}{x} ^ 2\right)} \mathrm{dx} ,$

$\therefore {I}_{1} = \int \frac{1 + \frac{1}{x} ^ 2}{{x}^{2} + \frac{1}{x} ^ 2} \mathrm{dx} .$

Subst., $x - \frac{1}{x} = u , \text{ so that, "(1+1/x^2)dx=du," and, also, }$

$\left({x}^{2} + \frac{1}{x} ^ 2\right) = {\left(x - \frac{1}{x}\right)}^{2} + 2 = {u}^{2} + 2.$

$\therefore {I}_{1} = \int \frac{1}{{u}^{2} + 2} \mathrm{du} = \frac{1}{\sqrt{2}} a r c \tan \left(\frac{u}{\sqrt{2}}\right) .$

As, $u = x - \frac{1}{x} = \frac{{x}^{2} - 1}{x} ,$ we have,

${I}_{1} = \frac{1}{\sqrt{2}} a r c \tan \left(\frac{{x}^{2} - 1}{\sqrt{2} \cdot x}\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(1\right) .$

Next, ${I}_{2} = \int \frac{{x}^{2} - 1}{{x}^{4} + 1} \mathrm{dx} = \int \frac{{x}^{2} \left(1 - \frac{1}{x} ^ 2\right)}{{x}^{2} \left({x}^{2} + \frac{1}{x} ^ 2\right)} \mathrm{dx} ,$

$\therefore {I}_{2} = \int \frac{{x}^{2} - \frac{1}{x} ^ 2}{{x}^{2} + \frac{1}{x} ^ 2} \mathrm{dx} .$

Take $x + \frac{1}{x} = v , \text{ so that, "(1-1/x^2)dx=dv," and, further, }$

${x}^{2} + \frac{1}{x} ^ 2 = {\left(x + \frac{1}{x}\right)}^{2} - 2 = {v}^{2} - 2.$

$\therefore {I}_{2} = \int \frac{1}{{v}^{2} - {\left(\sqrt{2}\right)}^{2}} \mathrm{dv} = \frac{1}{2 \sqrt{2}} \ln | \frac{v - \sqrt{2}}{v + \sqrt{2}} | .$

Replacing $v , b y , x + \frac{1}{x} = \frac{{x}^{2} + 1}{x} ,$ we get,

${I}_{2} = \frac{1}{2 \sqrt{2}} \ln | \frac{{x}^{2} - \sqrt{2} \cdot x + 1}{{x}^{2} - \sqrt{2} \cdot x + 1} | \ldots \ldots \left(2\right) .$

Finally, observe that, the Reqd. Integral $I ,$ is,

$I = \int \frac{1}{{x}^{4} + 1} \mathrm{dx} = \frac{1}{2} \int \frac{\left({x}^{2} + 1\right) - \left({x}^{2} - 1\right)}{{x}^{4} + 1} \mathrm{dx} ,$

$\Rightarrow I = \frac{1}{2} \cdot {I}_{1} - \frac{1}{2} \cdot {I}_{2} ,$

$= \frac{1}{2 \sqrt{2}} a r c \tan \left(\frac{{x}^{2} - 1}{\sqrt{2} \cdot x}\right) - \frac{1}{4 \sqrt{2}} \ln | \frac{{x}^{2} - \sqrt{2} \cdot x + 1}{{x}^{2} - \sqrt{2} \cdot x + 1} | + C .$

N.B. : $J = \int {x}^{2} / \left({x}^{4} + 1\right) \mathrm{dx} = \frac{1}{2} \cdot {I}_{1} + \frac{1}{2} \cdot {I}_{2.}$

Enjoy Maths.!