If `int_0^1 (e^t dt)/(t+1)=a,` then evaluate `int_(b-1)^b (e^t dt)/(t-b-1)`

If int_(0)^(1)(e^(t))/(t+1)dt=a, then int_(b-1)^(b)(e^(-t))/(t-b-1)dt=

If rArr int_(0)^(1) (e^(-t))/(t+1) dt =a, "then"int_(b-1)^(b) (e^(-1))/(t-b-1)dt is equal to

If int_(0)^(1)(e^(t))/(1+t)dt=a, then find the value of int_(0)^(1)(e^(t))/((1+t)^(2))dt in terms of a

Let A=int_(0)^(1)(e^(t))/(t+1)dt, then the value of (int_(0)^(1)te^(t^^2))/(t^(2)+1)dtA^(2)(b)(1)/(2)A(c)2A(d)(1)/(2)A^(2)

Let A = int_(0)^(1)(e^(t))/(1+t) dt , then int_(a-1)^(a)(e^(-1))/(t-a-1) dt has the value :

If k=int_(0)^(1) (e^(t))/(1+t)dt , then int_(0)^(1) e^(t)log_(e )(1+t)dt is equal to

int_(1)^(a)(ln t)/(t)dt