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Question

11e2xdx is equal to

A
xlog[1+1e2x]+C
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B
log (ex+1)xex+C
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C
log(e2x+1)x+c
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D
12(e2x+1)+c
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Solution

The correct option is A xlog[1+1e2x]+C
I=11e2xdx=exe2x1dx
Put ex=texdx=dt,
I=1t21dt
=log[t+t21]+C
log[ex+e2x1]+C
=log[1ex+1e2xex]+C
=log[1+1e2x]+log ex+C
=xlog[1+1e2x]+C

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