`int (1+3t)t^2dt`

int sqrt(t)dt

Let f(x) be a function defined by f(x)=int_1^xt(t^2-3t+2)dt,1<=x<=3 Then the range of f(x) is

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 .

int (3 t^(2) + 5) dt :

If int_0^xf(t) dt=x+int_x^1 tf(t)dt, then the value of f(1)

I=int sqrt(1-t^(2))dt

If [int_0^1(dt)/(t^2+2tcosalpha+1)]x^2-[int_- 3^3(t^2sin2t)/(t^2+1)dt]x-2=0 (0 < alpha < pi) then the value of x is

Let function F be defined as f(x)= int_1^x e^t/t dt x > 0 then the value of the integral int_1^1 e^t/(t+a) dt where a > 0 is

The function int_(-1)^(x)t(e^t-1)(t-1)(t-2)^3(t-3)^5dt has local minimum at x=

If int_(1)^(x) (dt)/(|t|sqrt(t^(2)-t)) = (pi)/(6) , then x can be equal to :