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Question

Solve:
ex(1+ex)(2+ex) dx

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Solution

I=exdx(1+ex)(2+ex)
put ex=t
ex dx=dt
=dt(1+t)(2+t)=dt[(2+t)(1+t)(1+t)(2+t)]
=dt[1(1+t)1(2+t)]
I=dtt+1dt(2+t)
I=ln|t+1|ln|t+2|+ln|c|
I=lnc(t+1t+2)c= constant
I=lnc(ex+1ex+2)


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