`int (dv)/(1-v^3)`

Evaluate int (dv)/(1+2v)

int((1+v)/(1-v))dv

A body moves so that it follows the following relation (dv)/(dt)=-v^(2)+2v-1 where v is speed in m/s and t is time in second. If at t=0, v=0 then find the speed (in m/s) when acceleration in one fourth of its initial value.

Evaluate : int (( 2v ) / (v^2 -1) ) dv

int(1)/sqrt(1-2v^(2))dv

int(2v^(2))/(v^(2)+v-2)backslash dv

The motion of a body falling from rest in a medium is given by equation, (dv)/(dt)=1-2v . The velocity at any time t is

The motion of a particle moving along x-axis is represented by the equation (dv)/(dt)=6-3v , where v is in m/s and t is in second. If the particle is at rest at t = 0 , then

If u=f(x^2), v=g(x^3) , f'(x)=sinx and g'(x)= cos x then find (du) / (dv) .

Integrate : int((1)/(2v+v^(2)))dv