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Question Number 150993 by talminator2856791 last updated on 17/Aug/21
∫01ln(x+1)x2+1dx=?
Commented by puissant last updated on 17/Aug/21
Q150986
Answered by Ar Brandon last updated on 17/Aug/21
∫01ln(x+1)x2+1dx=∫0π4ln(1+tanϑ)dϑ=∫0π4ln(sinϑ+cosϑ)dϑ−∫0π4ln(cosϑ)dϑ=12(G−πln24)−(G2−πln24)=−πln28+πln24=πln28
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