Question

integration of g'(x)÷(g(x))^2 dx=

Answer 1

it is given that $\int{\frac{g'(x)}{g(x)^2}}\,dx$

Let g(x) = P
differentiate with respect to x both sides,
g'(x) = dP/dx
so, g'(x).dx = dP ......(i)

now, $\int{\frac{g'(x)}{g(x)^2}}\,dx=\int{dP/P^2}$

= $\frac{P^{-2+1}}{-2+1}$ + C
= - $\frac{1}{P}$ + C

now, put P = g(x)

so, $\int{\frac{g'(x)}{g(x)^2}}\,dx$ = -1/g(x) + C