" If "u" and "v" are two differentiable functions of "x" then "int uvdx=u int vdx-int(((du)/(dx))*int vdx)dx

If u and v are two differentiable functions of x and int uv dx=phi(x)-int[(du)/(dx)intv dx]dx , then the value of phi(x) is-

If u and v are two functions of x then prove that: int uvdx=u int vdx-int[(du)/(dx)int vdx]dx

If u and v are two functions of x then prove that : int uv dx = u int v dx - int [ (du)/(dx) int v dx ] dx Hence, evaluate int x e^(x) dx

If u and v are integrable function of x, then, show that intu.v.dx=uintvdx-int[(du)/(dx)intvdx]dx . Hence evaluate intlogxdx .

d/(dx)(int f(x)dx)=

If u and v are two differentiable functions of x show that, (d)/(dx)("tan"^(-1)(u)/(v))=(v (du)/(dx)-u(dv)/(dx))/(u^(2)+v^(2)) .

If f is a continuous function on [a;b] and u(x) and v(x) are differentiable functions of x whose values lie in [a;b] then (d)/(dx){int_(u(x))^(v)(x)f(t)dt}=f(v(x))dv(x)/(dx)-f(u(x))du(x)/(dx)

If u, v, w are three differentiable functions of x , prove that d/dx (uvw) = ((du)/dx)vw + u ((dv)/dx) w + uv ((dw)/dx) . Use this result to differentiate x^(3)(x^(2)+1)(x^(4)+1) w.r.t. x.

If y = f(u) is a differentiable functions of u and u = g(x) is a differentiable functions of x such that the composite functions y = f[g(x) ] is a differentiable functions of x then (dy)/(dx) = (dy)/(du) . (du)/(dx) Hence find (dy)/(dx) if y = sin ^(2)x