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Question Number 154469 by mnjuly1970 last updated on 18/Sep/21
prove:∫01ln(4−2x+x2)dx=2ln(2e)+π3
Answered by ARUNG_Brandon_MBU last updated on 18/Sep/21
I=∫01ln(4−2x+x2)dx=∫01ln(3+(1−x)2)dx=∫01ln(3+x2)dx=[xln(3+x2)]01−∫012x2x2+3dx=ln4−2∫01(1−3x2+3)dx=2ln2−2+23[tan−1(x3)]01=2ln2−2lne+33π=2ln(2e)+π3
Commented by mnjuly1970 last updated on 18/Sep/21
grateful..verynicesolution..
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