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

Question

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

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Answer

let $\text{I}=\int\frac{\text{x}^2-3\text{x}+1}{\text{x}^4+\text{x}^2+1}\ \text{dx}$
Dividing numerator and denominator bt $x^2$
$\therefore\text{I}=\int\frac{1-\frac{3}{\text{x}}+\frac{1}{\text{x}^2}}{\text{x}^2+1+\frac{1}{\text{x}^2}}\ \text{dx}$
$=\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}}{\Big(\text{x}-\frac{1}{\text{X}}\Big)^2+3}-\int\frac{3\text{x}}{\text{x}^4+\text{x}^2+1}\ \text{dx}$
Let $\Big(\text{x}-\frac{1}{\text{x}}\Big)=\text{t}$
$\Rightarrow\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt} [$For $1^{st}$ part$]$
Let $\text{x}^2=\text{z}$
$\Rightarrow2\text{x dx}=\text{dz} [$For $2^{nd}$ part$]$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}^2+3}-\frac{3}{2}\int\frac{\text{dz}}{\text{z}^2+\text{z}+1}$
$\Rightarrow\text{I}=\int\frac{\text{dt}}{\text{t}^2+3}-\frac{3}{2}\int\frac{\text{dz}}{\Big(\text{z}+\frac{1}{2}\Big)^2+\frac{3}{4}}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{t}}{\sqrt{3}}\Big)-\frac{3}{2}\times\frac{2}{\sqrt{3}}\tan^{-1}\Bigg(\frac{\text{z}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\Bigg)+\text{C}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{x}^2-1}{\sqrt{3}\text{x}}\Big)-\sqrt{3}\tan^{-1}\Big(\frac{2\text{z}+1}{\sqrt{3}}\Big)+\text{C}$
Hence,
$\text{I}=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{x}^2-1}{\sqrt{3}\text{x}}\Big)-\sqrt{3}\tan^{-1}\Big(\frac{2\text{x}^2+1}{\sqrt{3}}\Big)+\text{C}$

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