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a
ex−12e2x−2ex+1+12logex−1+e2x−2ex+1+C
b
ex−12e2x−2ex−1+12logex−1+e2x−2ex−1+C
c
ex−12e2x−2ex−1+logex−1+e2x−2ex−1+C
d
ex−12e2x−2ex−1−logex−1+e2x−2ex−1+C
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Correct option is D
Let I=∫ex−e[x]−x−2e3x2dx,x∈(0,1)I=∫ex−e−x−2e3x2dxI=∫ex2−2ex−1e3x2dx Put ex=tI=∫t2−2t−1 dtI=∫(t−1)2−(2)2dt=(t−1)2(t−1)2−(2)2−(2)22log(t−1)+(t−1)2−(2)2+C=ex−12ex−12−2−logex−1+ex−12−2+C=ex−12e2x−2ex−1−logex−1+e2x−2ex−1+CSimilar Questions
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