Integrate x2/(x4+x2+1) - Maths -

Question

Integrate x2/(x4+x2+1)

Ajanta Trivedi · Accepted Answer

the given integral is:

$I = \int \frac{x^{2}}{x^{4} + x^{2} + 1} ⅆ x$ dividing numerator and denominator by $x^{2}$ .

$I = \int \frac{1}{x^{2} + 1 + 1 / x^{2}} ⅆ x = \frac{1}{2} \cdot \int \frac{(1 + \frac{1}{x^{2}}) + (1 - \frac{1}{x^{2}})}{x^{2} + \frac{1}{x^{2}} + 1} ⅆ x = \frac{1}{2} [\int \frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}} + 1} ⅆ x + \int \frac{1 - \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}} + 1} ⅆ x] = \frac{1}{2} [I_{1} + I_{2}]$

where

$I_{1} = \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}} + 1} ⅆ x = \int \frac{1 + \frac{1}{x^{2}}}{(x - \frac{1}{x})^{2} + 3} ⅆ x l e t x - \frac{1}{x} = t \Rightarrow (1 + \frac{1}{x^{2}}) d x = d t I_{1} = \int \frac{1}{t^{2} + (\sqrt{3})^{2}} ⅆ t I_{1} = \frac{1}{\sqrt{3}} \cdot \tan^{- 1} \frac{t}{\sqrt{3}} + C_{1} I_{1} = \frac{1}{\sqrt{3}} \cdot \tan^{- 1} [\frac{x - \frac{1}{x}}{\sqrt{3}}] + C_{1}$ where $C_{1}$ is an integration constant.

$I_{1} = \frac{1}{\sqrt{3}} \tan^{- 1} [\frac{x^{2} - 1}{\sqrt{3} x}] + C_{1}$

similarly

$I_{2} = \int \frac{1 - \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}} + 1} ⅆ x$

$I_{2} = \int \frac{1 - \frac{1}{x^{2}}}{(x + \frac{1}{x})^{2} - 1} ⅆ x l e t x + \frac{1}{x} = t \Rightarrow (1 - \frac{1}{x^{2}}) d x = d t$

$I_{2} = \int \frac{1}{t^{2} - 1} ⅆ t I_{2} = \frac{1}{2} \log \frac{t - 1}{t + 1} + C_{2}$ where $C_{2}$ is an integration constant.

$I_{2} = \frac{1}{2} \log \frac{x + \frac{1}{x} - 1}{x + \frac{1}{x} + 1} + C_{2} I_{2} = \frac{1}{2} \log \frac{x^{2} - x + 1}{x^{2} + x + 1} + C_{2}$

thus

$I = \frac{1}{2} [\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x^{2} - 1}{\sqrt{3} x}) + C_{1} + \frac{1}{2} \log \frac{x^{2} - x + 1}{x^{2} + x + 1} + C_{2}] = \frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{x^{2} - 1}{\sqrt{3} x}) + \frac{1}{4} \log \frac{x^{2} - x + 1}{x^{2} + x + 1} + C w h e r e C = \frac{C_{1}}{2} + \frac{C_{2}}{2}$

hope this helps you.

cheers!!