# How to integrate ln t/( 1-t)

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It is not ln (t/1-t) what I asked for the soluitions,The function needs to be integrated is [ ln t] /(1-t)

`intln(t/(1-t))dt=int(lnt-ln(1-t))dt`

`=intlntdt-intln(1-t)dt`

`` Integrating by parts

`int lnt dt=t lnt-t` and

`int ln(1-t) dt= t ln(1-t)+int (t/(1-t))dt`

`=t ln(1-t)+int(-(1-t)+1)/(1-t)dt`

`=t ln(1-t)-t-ln(1-t)`

Thus

`intln(t/(1-t))dt=tln(t)-t-tln(1-t)+t+ln(1-t)+c`

`=t(ln(t)-ln(1-t))+ln(1-t)+c`

Where c is integrating constant.