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$\int {\{\frac{(\log x - 1)}{1 + (\log x)^{2}}\}}^{2} dx$ is equal to

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x(log⁡x)2+1+C

xex1+x2+C

xx2+1+C

log⁡x(log⁡x)2+1+C

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detailed solution

Correct option is A

∫(log⁡x−1)1+(log⁡x)22dx=∫(log⁡x)2+1−2log⁡x(log⁡x)2+12dx=∫(log⁡x)2+1−2xlog⁡x⋅1x(log⁡x)2+12dx=∫ddxx(log⁡x)2+1dx=x(log⁡x)2+1+C

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∫(log⁡x−1)1+(log⁡x)22dx is equal to

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$\int {\{\frac{(\log x - 1)}{1 + (\log x)^{2}}\}}^{2} dx$ is equal to