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

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)/(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_0^1t e^t^2)/(t^2+1)dt A^2 (b) 1/2A (c) 2A (d) 1/2A^2

If lambda=int_0^1(e^t)/(1+t) , then int_0^1e^tlog_e(1+t)dt is equal to

int1/(sqrtt(1+t))dt

int(1)/(sqrt(t)+1)dt

int e^t sint dt

Suppose f(x) and g(x) are two continuous functions defined for 0<=x<=1 .Given, f(x)=int_0^1 e^(x+1) .f(t) dt and g(x)=int_0^1 e^(x+1) *g(t) dt+x The value of f( 1) equals

Let f(x)=(e^(x)+1)/(e^(x)-1) and int_(0)^(1) x^(3) .(e^(x)+1)/(e^(x)-1) dx= alpha "Then" , int_(-1)^(1) t^(3) f(t) dt is equal to

Let f(x) be a differentiable function such that f(x)=x^2 +int_0^x e^-t f(x-t) dt then int_0^1 f(x) dx=