What is the integration of #(log(1+e^x))/e^x# ?

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Answer 1 · 2016-09-07T07:27:06

Let #I=intlog(1+e^x)/e^xdx#

We will use the substn. #e^x=t", so that, "e^xdx=dt, or, dx=dt/e^x=dt/t#.

#:. I=intlog(1+t)/t*dt/t=intt^-2log(1+t)dt#

Now, we use the Rule of Integration by Parts (IBP) :

#"(IBP) : "intuvdt=uintvdt-int((du)/dtintvdt)dt#.

We take, #u=log(1+t) rArr (du)/(dt)=1/(1+t)#, and,

#v=t^-2 rArr intvdt=t^(-2+1)/(-2+1)=t^-1/-1=-1/t#.

#:. I=-1/tlog(1+t)-int{(1/(1+t))(-1/t)}dt#

#=-1/tlog(1+t)+int{1/((t(1+t))}dt#

#=-1/tlog(1+t)+int{(1+t-t)/((t(1+t))}dt#

#=-1/tlog(1+t)+int{(1+t)/(t(1+t))-t/(t(1+t))}dt#

#=-1/tlog(1+t)+int{1/t-1/(1+t)}dt#

#=-1/tlog(1+t)+lnt-ln(1+t)#

#=-1/e^xlog(1+e^x)+lne^x-ln(1+e^x)............".[as, "t=e^x]#

#=-log(1+e^x)/e^x+x-ln(1+e^x)#

#=x-(1/e^x+1)ln(1+e^x)+C#.

To be continued...