Basic Mathematics

Integration of
(Log X)^2/X
Let, Log X=t
d/dx (Log X)
=d/dx(t)
1/x=
d/dt(t)dx/dt
1/x=1(dx/dt )
(1/x)dx=dt
Integration of
{(LogX)^2}
(1/X) dx
= Int {(t)^2}dt
=t^3/3+c
(1/3)(LogX)^3+c

Cos(A+B)
=CosACosB
-SinASinB
For, A=B=x
Cos(x+x)
=CosxCosx
-SinxSinx
Or, Cos(2x)
=Cos²x-Sin²x
But,
Sin²x+Cos²x=1
Thus,
Sin²x=1-Cos²x
Now,
Cos2x=Cos²x
-(1-Cos²x)
=Cos²x
-1+Cos²x
Or,Cos2x
=2Cos²x-1
Hence,
Cos2x+1
=2Cos²x
Or,
2Cos²x
=1+Cos2x
Also,
Cos(2x)
=Cos²x-Sin²x
and,
Sin²x+Cos²x=1
Then,
Cos²x=1-Sin²x
Cos(2x)
=(1-Sin²x)
-Sin²x
=1-2Sin²x
Or, Cos(2x)
=1-2Sin²x
Or,
2Sin²x
=1-Cos(2x)
We have
(1+Cos2x)/
(1-Cos2x)
=2Cos²x/2Sin²x
=(Cosx/Sinx)²
=(Cotx)²
=(Cot²x)

x³-1=0
Or,
(x-1)(x²+x+1)=0
Thus,
x=1, or,
( x²+x+1)=0
Or,
x²+2x(1/2)
+(1/4) +1-(1/4)=0
(x+1/2)²+(3/4)=0
(x+1/2)²-(-3/4)=0
{(x+1/2)²-
(i_/3/2)²}=0
as, i=_/(-1)
{(x+1/2)+
(i_/3/2)}{(x+1/2)-
(i_/3/2)}=0
AS,
(a²-b²)=
(a+b)(a-b)
=0
Thus, either
(x+1/2)+
(i_/3/2)=0
Or,
(x+1/2)-
(i_/3/2)=0
Thus,either
x=
(-1-i_/3)/2
Or,
x=(-1+i_/3)/2