`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

int (du) / (u sqrt (u ^ (2) -1))

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

int(logx)^(2)dx=ux+c, then u=

int(vdu-udv)/(u^(2)+v^(2))=

int(udu+vdv)/(u^(2)+v^(2))=

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

Evaluate : (a) int_(u)^(upsilon)dv = int_(0)^(t)adt (b)int_(0)^(s)ds=int_(0)^(t)udt+int_(0)^(t)adt (c) int_(u)^(upsilong)udv=int_(0)^(s)ads

If u_(n)=int(log x)^(n)dx, then u_(n)+nu_(n-1) is equal to :

int(u(v(du)/(dx)-u(dv)/(dx)))/(v^(3))dx=