Evaluate the following integrals: x^2+1 x^4-x^2+1 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\text{x}^2+1}{\text{x}^4-\text{x}^2+1}\ \text{dx}$

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Answer

let $\text{I}=\int\frac{\text{x}^2+1}{\text{x}^4-\text{x}^2+1}\ \text{dx}$
Dividing numerator and denominator bt $x^2$
$\therefore\text{I}=\frac{\Big(1+\frac{1}{\text{x}^2}\Big)}{\text{x}^2-1+\frac{1}{\text{x}^2}}\ \text{dx}$
$=\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}}{\Big(\text{x}-\frac{1}{\text{x}}\Big)^2+1}$
let $\Big(\text{x}-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt}$
$\Rightarrow\text{I}=\int\frac{\text{dt}}{\text{t}^2+1}$
$=\tan^{-1}\text{t}+\text{C}$
$\therefore\text{I}=\tan^{-1}\Big(\frac{\text{x}^2-1}{\text{x}}\Big)+\text{C}$

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