# How will you integrate ? int(dx)/(1+x^4)^2

May 30, 2018

Factorize the denominator then apply partial fraction decomposition.

#### Explanation:

Let

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

Complete the square in the denominator:

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

Apply the difference of squares:

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

Apply partial fraction decomposition:

$I = \frac{1}{8 \sqrt{2}} \int \left\{\frac{2 x + \sqrt{2}}{{x}^{2} + \sqrt{2} x + 1} ^ 2 - \frac{2 x - \sqrt{2}}{{x}^{2} - \sqrt{2} x + 1} ^ 2 + \frac{3 \left(x + \sqrt{2}\right)}{{x}^{2} + \sqrt{2} x + 1} - \frac{3 \left(x - \sqrt{2}\right)}{{x}^{2} - \sqrt{2} x + 1}\right\} \mathrm{dx}$

Rearrange:

$I = \frac{1}{8 \sqrt{2}} \int \left\{\frac{2 x + \sqrt{2}}{{x}^{2} + \sqrt{2} x + 1} ^ 2 - \frac{2 x - \sqrt{2}}{{x}^{2} - \sqrt{2} x + 1} ^ 2 + \frac{3}{2} \frac{2 x + \sqrt{2}}{{x}^{2} + \sqrt{2} x + 1} - \frac{3}{2} \frac{2 x - \sqrt{2}}{{x}^{2} - \sqrt{2} x + 1} + \frac{3}{2} \frac{\sqrt{2}}{{x}^{2} + \sqrt{2} x + 1} + \frac{3}{2} \frac{\sqrt{2}}{{x}^{2} - \sqrt{2} x + 1}\right\} \mathrm{dx}$

Complete the square in the denominator of the last two terms:

$I = \frac{1}{8 \sqrt{2}} \int \left\{\frac{2 x + \sqrt{2}}{{x}^{2} + \sqrt{2} x + 1} ^ 2 - \frac{2 x - \sqrt{2}}{{x}^{2} - \sqrt{2} x + 1} ^ 2 + \frac{3}{2} \frac{2 x + \sqrt{2}}{{x}^{2} + \sqrt{2} x + 1} - \frac{3}{2} \frac{2 x - \sqrt{2}}{{x}^{2} - \sqrt{2} x + 1} + \frac{3 \sqrt{2}}{{\left(\sqrt{2} x + 1\right)}^{2} + 1} + \frac{3 \sqrt{2}}{{\left(\sqrt{2} x - 1\right)}^{2} + 1}\right\} \mathrm{dx}$

Integrate term by term:

$I = \frac{1}{8 \sqrt{2}} \left\{- \frac{1}{{x}^{2} + \sqrt{2} x + 1} + \frac{1}{{x}^{2} - \sqrt{2} x + 1} + \frac{3}{2} \ln | {x}^{2} + \sqrt{2} x + 1 | - \frac{3}{2} \ln | {x}^{2} - \sqrt{2} x + 1 | + 3 {\tan}^{-} 1 \left(\sqrt{2} x + 1\right) + 3 {\tan}^{-} 1 \left(\sqrt{2} x - 1\right)\right\}$

Simplify:

$I = \frac{1}{8 \sqrt{2}} \left\{\frac{2 \sqrt{2}}{{x}^{4} + 1} + \frac{3}{2} \ln | \frac{{x}^{2} + \sqrt{2} x + 1}{{x}^{2} - \sqrt{2} x + 1} | + 3 {\tan}^{-} 1 \left(\frac{\sqrt{2} x}{1 - {x}^{2}}\right)\right\}$