Evaluate the following integrals: 1 x^4+3x^2+1 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{1}{\text{x}^4+3\text{x}^2+1}\ \text{dx}$

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Answer

Let $\text{I}=\int\frac{1}{\text{x}^4+3\text{x}^2+1}\ \text{dx}$
Dividing numerator and denominator by $x^2$
$\therefore\text{I}=\int\frac{\frac{1}{\text{x}^2}}{\text{x}^2+3+\frac{1}{\text{x}^2}}\ \text{dx}$
$=\frac{1}{2}\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)-\Big(1-\frac{1}{\text{x}^2}\Big)}{\text{x}^2+3+\frac{1}{\text{x}^2}}\ \text{dx}$
$=\frac{1}{2}\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)}{\Big(\text{x}-\frac{1}{\text{x}}\Big)^2+5}\ \text{dx}-\frac{1}{2}\int\frac{\Big(1-\frac{1}{\text{x}^2}\Big)}{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2+1}$
Let $\Big(\text{x}-\frac{1}{\text{x}}\Big)=\text{t}$
$\Rightarrow\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt}$
And $\text{x}+\frac{1}{\text{x}}=\text{z}$
$\Rightarrow\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dz}$
$\therefore\text{}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2+5}-\frac{1}{2}\int\frac{\text{dz}}{\text{z}^2+1}$
$=\frac{1}{2\sqrt{5}}\tan^{-1}\Big(\frac{\text{t}}{\sqrt{5}}\Big)-\frac{1}{2}\tan^{-1}(\text{z})+\text{C}$
Hence,
$\text{I}=\frac{1}{2\sqrt{5}}\tan^{-1}\Big(\frac{\text{x}^2-1}{\sqrt{5}\text{x}}\Big)-\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}^2+1}{\text{x}}\Big)+\text{C}$

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