AP EAMCET · Maths · Definite Integration
\(\int_0^1 \frac{8 \log (1+x)}{1+x^2} d x=\)
- A \(\pi \log 2\)
- B \(\frac{\pi}{2} \log 2\)
- C \(\frac{\pi}{4} \log 2\)
- D \(\log 2\)
Answer & Solution
Correct Answer
(A) \(\pi \log 2\)
Step-by-step Solution
Detailed explanation
\(I=\int_0^1 \frac{8 \log (1+x)}{1+x^2} d x=8 \int_0^1 \frac{\log (1+x)}{1+x^2} \cdot d x\) Let \(x=\tan \theta d x=\sec ^2 \theta d \theta\) When \(x=0, \theta=0\) and when \(x=1, \theta=\frac{\pi}{4}\) So,…
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