Question

integration of one upon axe

Answer 1

Log axe plus c is required answer

Answer 2

You have

∫1xdx=ln|x|+C∫1xdx=ln⁡|x|+C

(Note that the "constant" CC might take different values for positive or negative xx. It is really a locally constant function.)

In the same way,

∫f′(x)f(x)dx=ln|f(x)|+C∫f′(x)f(x)dx=ln⁡|f(x)|+C

The last derivative is given by

ddx|f(x)|=sgn(f(x))f′(x)={f′(x)−f′(x) if f(x)>0  if f(x)<0


hope it help u..