Solving a problem of integral lnx dx
note : § as sign of integral
note : § as sign of integral
To solve the equation of y=§ lnx dx, we
use 2 theorems ( theorems(1) and theorems(2)) of::
-Theorem(1) § u dv = uv- §v du (partial integral )
-Theorem(2) § u dv = §u (dv/dx) dx
-Theorem(3) § uv dx = § u d(§ v dx)
-Theorem(2) § u dv = §u (dv/dx) dx
-Theorem(3) § uv dx = § u d(§ v dx)
We can simply apply the technique using the theorems above!
Example I.
y = § lnx dx
y = § lnx dx
step 1 . let u=lnx, v=x, following theorem(1)/partial integral, we obtain
§lnx dx = xlnx - § x d(lnx)
step 2. let u=x, v=lnx then dv/dx=1/x., following theorem(2) we obtain
= xlnx - § x (1/x) dx
step 3.
= xlnx - §1 dx
= xlnx - §1 dx
step 4.
= xlnx - x + c
= xlnx - x + c
Example II.
y = § x cosx dx
y = § x cosx dx
step 1. - following theorem(3), let u=x, v=cosx, § cosx dx=sinx., we obtain
§ x cosx dx = § x d(sinx)
step 2. - following theorem(1)/ partial integral, let u=x, v=sinx, we obtain
= xsinx- § sinx dx
step 3.
= xsinx- (-cosx) + c
= xsinx- (-cosx) + c
step 4.
= xsinx + cosx + c
= xsinx + cosx + c
Good luck students