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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int\frac{1}{\text{x}^4+\text{x}^2+1}\ \text{dx}$
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Answer

We have
$\text{I}=\int\frac{1}{\text{x}^4+\text{x}^2+1}\ \text{dx}$
$=\frac{1}{2}\int\frac{2\text{dx}}{\text{x}^4+\text{x}^2+1}$
$\Rightarrow\int\bigg(\frac{(\text{x}^2+1)-(\text{x}^2-1)}{\text{x}^4+\text{x}^2+1}\bigg)\text{dx}$
$\Rightarrow\frac{1}{2}\int\Big(\frac{\text{x}^2+1}{\text{x}^2+\text{x}^2+1}\Big)\text{dx}-\frac{1}{2}\int\Big(\frac{\text{x}^2-1}{\text{x}^4+\text{x}^2+1}\Big)\text{dx}$
$\Rightarrow\frac{1}{2}\int\Big(\frac{\text{x}^2+1}{\text{x}^4+\text{x}^2+1}\Big)\text{dx}-\frac{1}{2}\int\Big(\frac{\text{x}^2-1}{\text{x}^4+\text{x}^2+1}\Big)\text{dx}$
Dividing numerator and denominator bt $x^2​$​​​​​​
$\text{I}=\frac{1}{2}\int\Bigg(\frac{1+\frac{1}{\text{x}^2}}{\text{x}^2+1+\frac{1}{\text{x}^2}}\Bigg)\text{dx}-\frac{1}{2}\int\Bigg(\frac{1-\frac{1}{\text{x}^2}}{\text{x}^2+1+\frac{1}{\text{x}^2}}\Bigg)\text{dx}$
$=\frac{1}{2}\int\Bigg(\frac{1+\frac{1}{\text{x}^2}}{\text{x}^2+\frac{1}{\text{x}^2}-2+3}\Bigg)\text{dx}-\frac{1}{2}\int\frac{\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}}{\text{x}^2+\frac{1}{\text{x}^2}+2-1}$
$=\frac{1}{2}\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}}{\Big(\text{x}-\frac{1}{\text{x}}\Big)^2+(\sqrt{3})^2}-\frac{1}{2}\int\frac{\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}}{\Big(\text{x}+\frac{1}{\text{x}}\Big)-1^2}$
putting $\text{x}-\frac{1}{\text{x}}=\text{t}$
$\Rightarrow\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dp}$
$\therefore\text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2+(\sqrt{3})^2}-\frac{1}{2}\int\frac{\text{dp}}{\text{p}^2-1^2}$
$=\frac{1}{2}\times\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{t}}{\sqrt{3}}\Big)-\frac{1}{2}\times\frac{1}{2\times1}\Big|\frac{\text{p}-1}{\text{p}+1}\Big|+\text{C}$
$=\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{\text{x}-\frac{1}{\text{x}}}{\sqrt{3}}\Big)-\frac{1}{4}\log\Bigg|\frac{\text{x}+\frac{1}{\text{x}}-1}{\text{x}+\frac{1}{\text{x}}+1}\Bigg|+\text{C}$
$=\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{\text{x}^2-1}{\text{x}\sqrt{3}}\Big)-\frac{1}{4}\log\Big|\frac{\text{x}^2-\text{x}+1}{\text{x}^2+\text{x}+1}\Big|+\text{C}$

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