Integration of log(1+x)/1+x². Limits from 0 to 1. Answer is π/8*log2

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Neelam Papanai · Answer

Put x=tan(theta)So dx=sec²(theta)d(theta)Then when x=0,t=0When x=1 t=π/4Then apply P4Then open tan(π/4+ theta)Using identity tan(a+b). Then solve, nd take LCM then u will get the answer..

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Integration of log(1+x)/1+x². Limits from 0 to 1. Answer is π/8*log2

Integration of log(1+x)/1+x². Limits from 0 to 1. Answer is π/8*log2

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Integration of log(1+x)/1+x². Limits from 0 to 1. Answer is π/8*log2

Integration of log(1+x)/1+x². Limits from 0 to 1. Answer is π/8*log2

1 Answers

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