# How can you evaluate: int_0^2(1/(1+x^4))dx ?

Mar 24, 2015

First of all, let's try to factorize ${x}^{4} + 1$.

${x}^{4} + 1 = {x}^{4} + 1 + 2 {x}^{2} - 2 {x}^{2} = \left({x}^{4} + 2 {x}^{2} + 1\right) - 2 {x}^{2} =$

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

So, since these two quadratic polynomials can't be further factorized, we have to find $A , B , C , D$:

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

$= \frac{A x + B}{{x}^{2} - \sqrt{2} x + 1} + \frac{C x + D}{{x}^{2} + \sqrt{2} x + 1} =$

$= \frac{\left(A x + B\right) \left({x}^{2} + \sqrt{2} x + 1\right)}{{x}^{2} - \sqrt{2} x + 1} + \frac{\left(C x + D\right) \left({x}^{2} - \sqrt{2} x + 1\right)}{{x}^{2} + \sqrt{2} x + 1} =$

$\frac{A {x}^{3} + \sqrt{2} A {x}^{2} + A x + B {x}^{2} + \sqrt{2} B x + B + C {x}^{3} - \sqrt{2} C {x}^{2} + C x + D {x}^{2} - \sqrt{2} D x + D}{\left({x}^{2} - \sqrt{2} x + 1\right) \left({x}^{2} + \sqrt{2} x + 1\right)}$,

the two numerators are:

$1 = {x}^{3} \left(A + C\right) + {x}^{2} \left(\sqrt{2} A + B - \sqrt{2} C + D\right) + x \left(A + \sqrt{2} B + C - \sqrt{2} D\right) + B + D$

and, for the identity of the two polynomial:

$A + C = 0$

$\sqrt{2} A + B - \sqrt{2} C + D = 0$

$A + \sqrt{2} B + C - \sqrt{2} D = 0$

$B + D = 1$

So:

$A = - C$ and $B = 1 - D$ put in the second equation:

$\sqrt{2} \left(- C\right) + 1 - D - \sqrt{2} C + D = 0 \Rightarrow 2 \sqrt{2} C = 1 \Rightarrow$

$C = \frac{1}{2 \sqrt{2}} = \frac{1}{2 \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{4}$ and so $A = - \frac{\sqrt{2}}{4}$

and the third equation becomes:

$- \frac{\sqrt{2}}{4} + \sqrt{2} \left(1 - D\right) + \frac{\sqrt{2}}{4} - \sqrt{2} D = 0 \Rightarrow$

$\sqrt{2} - \sqrt{2} D - \sqrt{2} D = 0 \Rightarrow 2 \sqrt{2} D = \sqrt{2} \Rightarrow D = \frac{1}{2}$ and so: $B = \frac{1}{2}$.

Our integral becomes:

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

$= - \frac{\sqrt{2}}{4} \int \frac{x - \frac{1}{2} \cdot \frac{4}{\sqrt{2}}}{{x}^{2} - \sqrt{2} x + 1} \mathrm{dx} +$

$+ \frac{\sqrt{2}}{4} \int \frac{x + \frac{1}{2} \cdot \frac{4}{\sqrt{2}}}{{x}^{2} + \sqrt{2} x + 1} \mathrm{dx} =$

$= - \frac{\sqrt{2}}{4} \cdot \frac{1}{2} \int \frac{2 x - \frac{1}{2} \cdot \frac{4}{\sqrt{2}} \cdot 2}{{x}^{2} - \sqrt{2} x + 1} \mathrm{dx} +$

$+ \frac{\sqrt{2}}{4} \cdot \frac{1}{2} \int \frac{2 x + \frac{1}{2} \cdot \frac{4}{\sqrt{2}} \cdot 2}{{x}^{2} + \sqrt{2} x + 1} \mathrm{dx} =$

$= - \frac{\sqrt{2}}{8} \int \frac{2 x - \frac{4}{\sqrt{2}}}{{x}^{2} - \sqrt{2} x + 1} \mathrm{dx} +$

$+ \frac{\sqrt{2}}{8} \int \frac{2 x + \frac{4}{\sqrt{2}}}{{x}^{2} + \sqrt{2} x + 1} \mathrm{dx} =$

$= - \frac{\sqrt{2}}{8} \int \frac{2 x - \sqrt{2} + \sqrt{2} - \frac{4}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}}{{x}^{2} - \sqrt{2} x + 1} \mathrm{dx} +$

$+ \frac{\sqrt{2}}{8} \int \frac{2 x + \sqrt{2} - \sqrt{2} + \frac{4}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}}{{x}^{2} + \sqrt{2} x + 1} \mathrm{dx} =$

$= - \frac{\sqrt{2}}{8} \int \frac{2 x - \sqrt{2} + \sqrt{2} - 2 \sqrt{2}}{{x}^{2} - \sqrt{2} x + 1} \mathrm{dx} +$

$+ \frac{\sqrt{2}}{8} \int \frac{2 x + \sqrt{2} - \sqrt{2} + 2 \sqrt{2}}{{x}^{2} + \sqrt{2} x + 1} \mathrm{dx} =$

$= - \frac{\sqrt{2}}{8} \int \frac{2 x - \sqrt{2}}{{x}^{2} - \sqrt{2} x + 1} \mathrm{dx} +$

$- \frac{\sqrt{2}}{8} \cdot \left(- \sqrt{2}\right) \int \frac{1}{{x}^{2} - \sqrt{2} x + 1} \mathrm{dx} +$

$+ \frac{\sqrt{2}}{8} \int \frac{2 x + \sqrt{2}}{{x}^{2} + \sqrt{2} x + 1} \mathrm{dx} +$

$+ \frac{\sqrt{2}}{8} \cdot \left(\sqrt{2}\right) \int \frac{1}{{x}^{2} + \sqrt{2} x + 1} \mathrm{dx}$.

The first and the third of these last four integrals are easy, remembering that:

$\int \frac{f ' \left(x\right)}{f} \left(x\right) \mathrm{dx} = \ln | f \left(x\right) | + c$

So:

$\left(1\right) = - \frac{\sqrt{2}}{8} \ln \left({x}^{2} - \sqrt{2} x + 1\right) + c$ and

$\left(3\right) = \frac{\sqrt{2}}{8} \ln \left({x}^{2} + \sqrt{2} x + 1\right) + c$.

For the second integral:

$\left(2\right) = \frac{\sqrt{2}}{8} \cdot \sqrt{2} \int \frac{1}{{x}^{2} - \sqrt{2} x + 1} \mathrm{dx}$

let's take the denominator:

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

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

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

And so, remembering that:

$\int \frac{f ' \left(x\right)}{1 + {\left[f \left(x\right)\right]}^{2}} \mathrm{dx} = \arctan f \left(x\right) + c$

$\left(2\right) = \frac{\sqrt{2}}{8} \int \frac{\sqrt{2}}{\frac{1}{2} \left[{\left(\sqrt{2} x - 1\right)}^{2} + 1\right]} \mathrm{dx} =$

$= \frac{\sqrt{2}}{8} \cdot 2 \arctan \left(\sqrt{2} x - 1\right) + c = \frac{\sqrt{2}}{4} \arctan \left(\sqrt{2} x - 1\right) + c$.

With similar counts the fourth integral becomes:

$\left(4\right) = \frac{\sqrt{2}}{4} \arctan \left(\sqrt{2} x + 1\right) + c$.

NOW the last thing to do is to calcultate the defined integral:

${\int}_{0}^{2} \frac{1}{1 + {x}^{4}} \mathrm{dx} =$

$= {\left[- \frac{\sqrt{2}}{8} \ln \left({x}^{2} - \sqrt{2} x + 1\right) + \frac{\sqrt{2}}{4} \arctan \left(\sqrt{2} x - 1\right) + \frac{\sqrt{2}}{8} \ln \left({x}^{2} + \sqrt{2} x + 1\right) + \frac{\sqrt{2}}{4} \arctan \left(\sqrt{2} x + 1\right)\right]}_{0}^{2} =$

$= - \frac{\sqrt{2}}{8} \ln \left(5 - 2 \sqrt{2}\right) + \frac{\sqrt{2}}{4} \arctan \left(2 \sqrt{2} - 1\right) + \frac{\sqrt{2}}{8} \ln \left(5 + 2 \sqrt{2}\right) + \frac{\sqrt{2}}{4} \arctan \left(2 \sqrt{2} + 1\right) +$

$- \left(- \frac{\sqrt{2}}{8} \ln 1 + \frac{\sqrt{2}}{4} \arctan \left(- 1\right) + \frac{\sqrt{2}}{8} \ln 1 + \frac{\sqrt{2}}{4} \arctan \left(1\right)\right) =$

$= - \frac{\sqrt{2}}{8} \ln \left(5 - 2 \sqrt{2}\right) + \frac{\sqrt{2}}{4} \arctan \left(2 \sqrt{2} - 1\right) + \frac{\sqrt{2}}{8} \ln \left(5 + 2 \sqrt{2}\right) + \frac{\sqrt{2}}{4} \arctan \left(2 \sqrt{2} + 1\right) - \frac{\sqrt{2}}{4} \left(- \frac{\pi}{4}\right) - \frac{\sqrt{2}}{4} \frac{\pi}{4}$.

I hope that my passages are clear and correct! If there are questions...I am here!