`int (-1)/(t-1) dt - int 1/t dt`

int t/(t+b^2) dt

int_e^(e(-1)) 1/(t(t+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

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

Integrate the following : (i) int(t-(1)/(t))^(2)" dt " (ii) intsin(10t-50)" dt " (iii) inte^((100t+6))" dt "

f(x)=x^(3)+1-2x int_(0)^(x)g(t)dt and g(x)=x-int_(0)^(1)f(t)dt. Then yhe value of int_(0)^(1)f(t)dt lies in the interval

int log (5+t) dt

int(t^(2)+1)/(t^(4))dt