d/dx ( log x) = 1/x
X(logX-1) + C
The integral of f'(x) = 1 is f(x) = x + c
âˆ« d/dx f(x) dx = f(x) + C C is the constant of integration.
logx^2=2 2logx=2 logx=1 10^1=x x=10
-logx=21.1 logx=-21.1 e^-21.1=x
âˆ« sin(x) dx = -cos(x) + CC is the constant of integration.
âˆ« cos(x) dx = sin(x) + CC is the constant of integration.
y=logx y=10 logx= 10 10logx = 10log1 logx = log1 x = 1 //NajN
*First off if we assume this log to be base 10 next we can use the product rule (d/dx (3)*logx+d/dx(logx)*3) 1.derivative of a constant is zero so that gives us 0*logx as our first term (simplifies to zero) next we have to differentiate logx that gives us 3*(1/xln(10)) so that leaves 0logx+3*(1/xln(10)) simplify...... 3/xln(10)
âˆ« cot(x) dx = ln(sin(x)) + CC is the constant of integration.
âˆ« ex dx = ex + CC is the constant of integration.