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Question

ex/2sec2x(1+4tan2x)dx is equal to

A
4ex/2sec2x+c
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B
2ex/2sec2x+c
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C
ex/2sec2x+c
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D
12ex/2sec2x+c
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Solution

The correct option is C 2ex/2sec2x+c
ex/2sec2x(1+4 tan 2x)dx
ex/22(sec2x+4 tan 2x.sec2x)dx
It is in the form of
ex(f(x)+f1(x))dx
=exf(x)+c
=2ex/2sec2x+c
exsec2x=2ex/2sec2x+c

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