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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 C2ex/2sec2x+c ∫ex/2sec2x(1+4tan2x)dx ∫ex/22(sec2x+4tan2x.sec2x)dx It is in the form of ∫ex(f(x)+f1(x))dx =exf(x)+c =2ex/2sec2x+c ∫exsec2x=2ex/2sec2x+c