∫xdx/(1-5x)=(-1/5)∫[1+(1/5)/(x-1/5)]dx=(-1/5)[x+(1/5)ln|x-1/5|]+c.
∫[x/(1-5x)]dx=-∫[x/(5x-1)]dx=-∫[(1/5)+1/5(5x-1)]dx=-(1/5)x-(1/25)∫[1/(5x-1)]d(5x-1)=-(1/5)x-(1/25)ln∣5x-1∣+C;
