Evaluate `int(dx)/((e^(x)-1)).`

Explanation

Putting `e^(x)=t and e^(x)dx=dt,i.e., dx=(1)/(t),`we get
`I=int(dx)/((e^(x)-1))=int(dt)/(t(t-1)).`
`Let(1)/(t(t-1))=(A)/(t)+(B)/(t(t-1)).`
`Then,1-=A(t-1)+Bt.`
Putting `t=0`in (i) ,we get `A=-1.`
Putting `t=1`in (i) , we get `B=1.`
`therefore (1)/(t(t-1))=(-1)/(t)+(1)/((t-1)).`
`"Hence",I=int(dx)/((e^(x)-1))`
`=int(dt)/(t(t-1))=int(-1)/(t)dt+int(1)/((t-1))dt`
`=-log |t|+log|t-1|+C`
`=log |(t-1)/(t)|+C`
`=log |(e^(x)-1)/(e^(x))|+C`