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Indefinite Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIndefinite Integrals5 Marks
Question
Evaluate the following integrals: $\int\frac{(\text{x}-1)^2}{\text{x}^4+\text{x}^2+1}\text{ dx}$
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Answer

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

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