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Indefinite Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int\frac{\text{x}^2+1}{\text{x}^2+7\text{x}^2+1}\ \text{dx}$
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Answer

We have,
$\text{I}=\int\Big(\frac{\text{x}^2+1}{\text{x}^4+7\text{x}^2+1}\Big)\text{dx}$
Dividing numerator and denominator by $x^2$
$\text{I}=\int\Bigg(\frac{1+\frac{1}{\text{x}^2}}{\text{x}^2+7+\frac{1}{\text{x}^2}}\Bigg)\text{dx}$
$=\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}}{\text{x}^2+\frac{1}{\text{x}^2}-2+9}$
$\Rightarrow\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}}{\Big(\text{x}-\frac{1}{\text{x}^2}\Big)^2+3^2}$
Putting $\text{x}-\frac{1}{\text{x}}=\text{t}$
$\Rightarrow\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}^2+3^2}$
$=\frac{1}{3}\tan^{-1}\Big(\frac{\text{t}}{3}\Big)+\text{C}$
$=\frac{1}{3}\tan^{-1}\Big(\frac{\text{x}-\frac{1}{\text{x}}}{3}\Big)+\text{C}$
$=\frac{1}{3}\tan^{-1}\Big(\frac{\text{x}^2-1}{3\text{x}}\Big)+\text{C}$

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