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Question

Evaluate:

dxex+2ex3dx.

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Solution

I=dxex+2ex3=exdxe2x3ex+2
ex=texdx=dt
I=dtt23t+2=dtt22tt+2=dtt(t2)1(t2)
=dt(t1)(t2)=dt[1(t2)1(t1)]
I=dt(t2)dt(t1)
I=ln|t2|ln|t1|+ln|c|
I=lnc(t2)t1=ln(ex2)(ex1)

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