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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
Choose the correct option from given four options$:\ \int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{dx}$ is equal to:
  • A
    $\frac{1}{5\text{x}}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
  • B
    $\frac{1}{5}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
  • C
    $\frac{1}{10\text{x}}(1+4)^{-5}+\text{C}$
  • ✓
    $\frac{1}{10}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$
✓

Answer

Correct option: D.
$\frac{1}{10}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$
Let $​\text{I}=\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{dx}=\int\frac{\text{x}^9}{\text{x}^{12}\Big(4+\frac{1}{\text{x}^2}\Big)^6}\text{dx}=\int\frac{\text{dx}}{\text{x}^3\Big(4+\frac{1}{\text{x}^2}\Big)^6}$
Put $4+\frac{1}{\text{x}^2}=\text{t}$
$\Rightarrow\frac{-2}{\text{x}^3}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^6}=-\frac{1}{2}\Big[\frac{\text{t}^{-6+1}}{-6+1}\Big]+\text{C}$
$=\frac{1}{10}\Big[\frac{1}{\text{t}^5}\Big]+\text{C}=\frac{1}{10}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$

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