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Question

exe2x+5ex+6dx=

A
log|ex+2ex+3|+c
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B
log|ex+3ex+2|+c
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C
log|ex2ex3|+c
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D
log|ex3ex2|+c
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Solution

The correct option is A log|ex+2ex+3|+c

ex=t

exdx=dt

dtt2+5t+6=dt(t+3)(t+2)

=[1(t+2)1(t+3)]dt

=1(t+2)dt1(t+3)dt

=log(t+2)log(t+3)+c

=log(t+2)(t+3)+c

=log(ex+2)(ex+3)+c


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