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Question Number 137940 by benjo_mathlover last updated on 08/Apr/21
∫1e1+1x2+1(x+1)2dx=?
Answered by john_santu last updated on 08/Apr/21
J=∫1ex2+1x2+1(x+1)2dxJ=∫1e(x2+1)(x+1)2+x2x2(x+1)2dxJ=∫1e(x2+x+1)2x2(x+1)2dxJ=∫1ex2+x+1x(x+1)dxJ=∫1e(1+1x−1x+1)dx=[x+ln∣x∣−ln∣x+1∣]1e=(e+1−ln(e+1))−(1−ln2)=e+ln(2e+1)
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