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 Evaluate ex4e2xdx

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a
12sin−1⁡ex2+C
b
sin−1⁡ex2+C
c
sin−1⁡ex2+C
d
None of these

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detailed solution

Correct option is B

I=∫ex4−e2xdx=∫ex22−ex2dx Let ex=t or exdx=dt∴ I=∫dt4−t2=∫dt22−t2 =sin−1⁡t2+C=sin−1⁡ex2+C


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