[" Q.32) If "u" and "v" are two functions of "x],[" ,then prove that "^(*)],[int u.vdx=u int vdx-int((d)/(dx)(u)*int vdx)dx]

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 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, v and w are functions of x, then show that (d)/(dx) (u.v.w) = (du)/(dx) v.w + u.(dv)/(dx).w + u.v.(dw)/(dx) in two ways- first by repeated application of product rule, second by logarithmic differentiation. Using product rule

If u, v and w are functions of x, then show that d/(dx)(udotvdotw)=(d u)/(dx)vdotw+udot(d v)/(dx)dotw+udotv(d w)/(dx) in two ways - first by repeated application of product rule, second by logarithmic differentiation.

If u, v and w are functions of x, then show that d/(dx)(u.v.w)=(d u)/(dx)v.w+u.(d v)/(dx).w+u.v(d w)/(dx) in two ways - first by repeated application of product rule, second by logarithmic differentiation.

If u, v and w are functions of x, then show that (d)/(dx)(u.v.w) = (du)/(dx) v.w+u. (dv)/(dx).w+u.v(dw)/(dx) in two ways-first by repeated application of product rule, second by logarithmic differentiation.

If u, v and w are functions of x, then show that (d)/(dx)(u.v.w) = (du)/(dx) v.w+u. (dv)/(dx).w+u.v(dw)/(dx) in two ways-first by repeated application of product rule, second by logarithmic differentiation.