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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
Integration of $\frac{1}{1+(\log_\text{e}\text{x})^2}$ with respect to $\log_\text{e}\text{x}$ is:
  • A
    $\frac{\tan^{-1}(\log_\text{e}\text{x})}{\text{x}}+\text{C}$
  • ✓
    $\tan^{-1}(\log_\text{e}\text{x})+\text{C}$
  • C
    $\frac{\tan^{-1}\text{x}}{\text{x}}+\text{C}$
  • D
    none of these
✓

Answer

Correct option: B.
$\tan^{-1}(\log_\text{e}\text{x})+\text{C}$
$\frac{1}{1+(\log_\text{e}\text{x})^2}\text{ d}(\log_\text{e}\text{x})$
Put $\log_\text{e}\text{x}=\text{t}$
$\int\frac{\text{dt}}{1+\text{t}^2}=\tan^{-1}\text{t}+\text{C}$
$\tan^{-1}(\log_\text{e}\text{x})+\text{C}$

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