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Question Number 125194 by bemath last updated on 09/Dec/20
∫(1−x2+x+1)2x2x2+x+1dx?
Answered by liberty last updated on 09/Dec/20
bysecondEulersubstitutionwesetx2+x+1=xt+1;thenx2+x+1=x2t2+2xt+1;x=2t−11−t2dx=2t2−2t+2(1−t2)2dt;I=∫(1−x2+x+1)2x2x2+x+1dx=2∫t21−t2dtI=−2t+ln∣1+t1−t∣+cI=−2(x2+x+1−1)x+ln∣x+x2+x+1−1x−x2+x+1+1∣+c=−2(x2+x+1−1)x+ln∣2x+2x2+x+1+1∣+c
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