AP EAMCET · Maths · Definite Integration
\(\int_0^1 \frac{\log _e(1+x)}{1+x^2} d x=\)
- A \(\frac{\pi}{4} \log _6 2\)
- B \(\frac{\pi}{6} \log _6 6\)
- C \(\frac{\pi}{2} \log _6 8\)
- D \(\frac{\pi}{8} \log _6 2\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi}{8} \log _6 2\)
Step-by-step Solution
Detailed explanation
Given integral \(I=\int_0^1 \frac{\log _e(1+x)}{1+x^2} d x\) put \(x=\tan \theta\), so at \(x=0, \theta=0\) and at \(x=1, \theta=\frac{\pi}{4}\) and \(d x=\sec ^2 \theta d \theta\) Now,…
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