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Question

e3x+exe4xe2x+1dx=

A
14log(e4xe2x+1)+c
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B
tan1(exex)+c
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C
tan1(ex+ex)+c
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D
tan1(exex)+c
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Solution

The correct option is B tan1(exex)+c

[e3x+exe4xe2x+1]dx

e2x(e+1ex)(e4xe2x+1)dx=[ex+1exe2x+1e2x+1]dx

ex1ex=t(ex+1ex)dx=dt

dtt2+1=dtt2+1

11tan1(t1)+c

tan1(exex)+c


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