Evaluate `I=int u/(a^4+u^4)du`

Evaluate int (du)/(1+sinu+cosu)

int u logudu

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

Find the value of int_(u)^(v)Mvdv

int(1)/(sqrt(u))du

Suppose an object has thickness du so that it extends from object distance u to u+du. Prove that the thickness dv of its image is given by (-v^2/u^2)du, so the longitudinal magnification (dv)/(du)=-m^2, where m is the lateral magnification.

int_(1)^(2)((x^(2)-1)dx)/(x^(3)*sqrt(2x^(4)-2x^(2)+1))=(u)/(v) where u are ( in their blowest form.Find the value of )/(sqrt(2x^(4)-2x^(2)+1))=(u)/(v) where of ((1000)u)/(v)

The function u=e^(x)sin x,v=e^(x)cos x satisfy the equation (a) v(du)/(dx)-u(du)/(dx)=u^(2)+v^(2)(b)(d^(2)u)/(dx^(2))=2v(c)d^(2)v())/(dx^(2))=-2u(d)(du)/(dx)+(dv)/(dx)=2v