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Indefinite Integrals — Maths STD 12 Science — Question

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

Let $\text{I}=\int\frac{\text{x}^2(\text{x}^4+4)}{\text{x}^2+4}\text{ dx}$
$=\int\frac{\text{x}^6+4\text{x}^2}{(\text{x}^2+4)}\text{ dx}$
$=\int\Big[\text{x}^4-4\text{x}^2+20-\frac{80}{\text{x}^2+4}\Big]\text{dx}$
$\text{I}=\frac{\text{x}^5}{5}-\frac{4\text{x}^3}{3}+20\text{x}-80\int\frac{1}{\text{x}^2+4}\text{ dx}+\text{C}_1\ ....(1)$
Let $\text{I}_1=\int\frac{1}{\text{x}^2+4}\text{ dx}$
$\text{I}_1=\int\frac{1}{\text{x}^2+(2)^2}\text{ dx}$
$\text{I}_1=\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}_2\ ....(2)$ $\Big[\text{Since},\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
Using equation (1) and (2)
$\text{I}=\frac{\text{x}^5}{5}-\frac{4\text{x}^3}{3}+20\text{x}-\frac{80}{2}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$
$\text{I}=\frac{\text{x}^5}{5}-\frac{4\text{x}^3}{3}+20\text{x}-40\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$

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