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This question already has an answer here:

$$\int_{0}^{1} \frac{\log{(1+x)}}{x^{2}+1} \ dx $$

I tried substituting x with 1/t but couldn't find the answer. Can someone provide any hint?

As many suggested I substituted x with tan t but again I got stuck at $$\int_{0}^{\frac{π}{4}} \log{(1+tant)} \ dt $$

I finally solved it and got the correct answer.

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marked as duplicate by Renascence_5., miracle173, Claude Leibovici, zhoraster, Watson Jan 21 '17 at 10:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ $x=\tan \theta$ $\endgroup$ – Anurag A Jan 21 '17 at 7:14
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Hint -

Put $\tan^{-1} x = t$

Then x = tan t.

And $\frac{1}{1 + x^2} dx = dt$

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  • $\begingroup$ All downvotes are welcome. But please tell the reason so that I can improve my mistake. $\endgroup$ – Kanwaljit Singh Jan 21 '17 at 10:49
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Put $\displaystyle x = \frac{1-t}{1+t}$

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