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ex2e2x+3ex+2dx is equal to 

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a
12logex+1ex+2+k
b
12logex−1ex+1+k
c
logex+1ex+2+k
d
logex+2ex+1+k

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detailed solution

Correct option is C

∫ex2e2x+3ex+2dx Put ex=texdx−dt∫exex2+3ex+2dx=∫dtt2+3t+2=∫dtt2+232t+322−322+2 Use ∫1x2−a2dx=12alog⁡x−ax+a+c=∫dtt+322−122=12⋅12log⁡t+32−12t+32+12+k=log⁡2ex+22ex+4+k=log⁡ex+1ex+2+k


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