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$\int_{0}^{π / 4} \log (1 + \tan^{2} θ + 2 \tan θ) d θ =$

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detailed solution

Correct option is C

Let I=∫0π/4 log⁡(1+tan⁡θ)dθ=∫0π/4 log⁡(1+tan⁡(π/4−θ))dθ=∫0π/4 log⁡1+1−tan⁡θ1+tan⁡θdθ=∫0π/4 [log⁡2−log⁡(1+tan⁡θ)]dθ⇒ 2I=π/4log⁡2.Required integral =2I=π4log⁡2

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∫0π/4 log⁡1+tan2⁡θ+2tan⁡θdθ=

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$\int_{0}^{π / 4} \log (1 + \tan^{2} θ + 2 \tan θ) d θ =$