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

$\text{I}=\int\frac{\text{x}^2+1}{\text{x}^4+\text{x}^2+1}\ \text{dx}$
$=\int\frac{1+\frac{1}{\text{x}^2}}{\text{x}^2+1+\frac{1}{\text{x}^2}}\ \text{dx}$
Dividing numerator and denominator by $x^2$
$=\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)}{\Big(\text{x}-\frac{1}{\text{x}}\Big)+3}\ \text{dx}$
Let $\text{x}-\frac{1}{\text{x}}=\text{t}\Rightarrow\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt}$
$\therefore\text{I}=\frac{\text{dt}}{\text{t}^2+3}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{t}}{\sqrt{3}}\Big)+\text{C}$
$\therefore\text{I}=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{x}^2-1}{\sqrt{3}\text{x}}\Big)+\text{C}$

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