Integral of $\frac{1}{1+(\log x)^2}$ w.r

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Indefinite Integration

Question:

Integral of $\frac{1}{1+(\log x)^2}$ w.r.t. log x is:

Options:

$\frac{\tan^{-1}(\log x)}{x}+C$

$\tan^{-1}(\log x)+C$

$\frac{\tan^{-1}x}{x}+C$

None of these

Correct Answer:

$\tan^{-1}(\log x)+C$

Explanation:

$\frac{1}{1+(\log x)^2}d(\log x)=\int\frac{dt}{1-t^2}$ [Putting log x = t ⇒ $d(\log x)=dt=\tan^{-1}t+C=\tan^{-1}(\log x)+C$