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

Jul 9, 2017

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

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

= $\frac{1}{4} \cdot \int {x}^{- 3} \cdot 4 {x}^{- 5} \cdot \frac{\mathrm{dx}}{{x}^{- 4} + 1} ^ 2$

= $\frac{1}{4} \cdot {x}^{- 3} \cdot {\left({x}^{- 4} + 1\right)}^{- 1} - \frac{1}{4} \cdot \int \left(- 3\right) \cdot {x}^{- 4} \cdot \frac{\mathrm{dx}}{{x}^{- 4} + 1}$

= $\frac{1}{4} \cdot {x}^{- 3} / \left({x}^{- 4} + 1\right) + \frac{3}{4} \cdot \int {x}^{- 4} \cdot \frac{\mathrm{dx}}{{x}^{- 4} + 1}$

= $\frac{1}{4} \cdot \frac{x}{{x}^{4} + 1} + \frac{3}{4} \cdot \int \frac{\mathrm{dx}}{{x}^{4} + 1}$

= $\frac{1}{4} \cdot \frac{x}{{x}^{4} + 1} + \frac{3}{8} \cdot \int 2 \cdot \frac{\mathrm{dx}}{{x}^{4} + 1}$

= $\frac{1}{4} \cdot \frac{x}{{x}^{4} + 1} + \frac{3}{8} \cdot \int 2 {x}^{- 2} \cdot \frac{\mathrm{dx}}{{x}^{2} + {x}^{- 2}}$

= $\frac{1}{4} \cdot \frac{x}{{x}^{4} + 1} + \frac{3}{8} \cdot \int \left(1 + {x}^{- 2}\right) \cdot \frac{\mathrm{dx}}{{x}^{2} + {x}^{- 2}} - \frac{3}{8} \cdot \int \left(1 - {x}^{- 2}\right) \cdot \frac{\mathrm{dx}}{{x}^{2} + {x}^{- 2}}$

= $\frac{1}{4} \cdot \frac{x}{{x}^{4} + 1} + \frac{3}{8} \cdot \int \left(1 + {x}^{- 2}\right) \cdot \frac{\mathrm{dx}}{{\left(x - {x}^{- 1}\right)}^{2} + 2} - \frac{3}{8} \cdot \int \left(1 - {x}^{- 2}\right) \cdot \frac{\mathrm{dx}}{{\left(x + {x}^{- 1}\right)}^{2} - 2}$

= $\frac{1}{4} \cdot \frac{x}{{x}^{4} + 1} + \frac{3}{16} \cdot S q r t \left(2\right) \cdot A r c \tan \left[\frac{x - {x}^{- 1}}{2} ^ \left(\frac{1}{2}\right)\right] - \frac{3}{16} \cdot S q r t \left(2\right) \cdot A r c \sinh \left[\frac{x + {x}^{- 1}}{2} ^ \left(\frac{1}{2}\right)\right] + C$

= $\frac{1}{4} \cdot \frac{x}{{x}^{4} + 1} + \frac{3}{16} \cdot S q r t \left(2\right) \cdot A r c \tan \left[\frac{{x}^{2} - 1}{x \cdot S q r t \left(2\right)}\right] - \frac{3}{16} \cdot S q r t \left(2\right) \cdot A r c \sinh \left[\frac{{x}^{2} + 1}{x \cdot S q r t \left(2\right)}\right] + C$

#### Explanation:

1) I divided denominator and numerator of integrand with x^8

2) I decomposed numerator for resembling derivative of denominator.

3) I used partial fraction

4) I expanded fractions with x^4

5) I started to decompose second integral by multiply and divide with 2

6) I divided denominator and numerator of second integrand with x^2

7) I decomposed second one for resembling denominators of them at forms of ${u}^{2} + {a}^{2}$ for u=$x - {x}^{- 1}$ and ${u}^{2} - {a}^{2}$ for u=$x + {x}^{- 1}$

8) I integrated decomposed them.

9) I rewrote results.