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Question

20 e2re2xe2ezx

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Solution

The integral of the function is given as,

e 2x e 2x e 2x + e 2x dx(1)

Consider, e 2x + e 2x =t.

Differentiate with respect to x.

e 2x + e 2x =t ( 2 e 2x 2 e 2x )dx=dt 2( e 2x e 2x )dx=dt

Substitute 2( e 2x e 2x )dx=dt in equation (1) and then integrate.

e 2x e 2x e 2x + e 2x dx= dt 2t = 1 2 1 t dt = 1 2 log| t |+C

Substitute e 2x + e 2x =t in the above equation.

e 2x e 2x e 2x + e 2x dx= 1 2 log| t |+C = 1 2 log| e 2x + e 2x |+C


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