If ∫(x+1)dxx(1+xex)2=log f(x)+11+xex+c , then f(x)=

I=∫(x+1)dxx(1+xex)2 Putting (1+xex)=t  ⇒  (x+1)exdx=dt I=∫dt(t−1)t2=∫(–1t2–1t+1t–1)dt    =1t–logt+log(t–1)+c    =11+xex+log(xex1+xex)+c ⇒ f(x)=xex1+xex .