Question

If f'(x)=x^{2}- 1/ x^{2} and f(1)=1/3 then find f(x)

Answer 1

Step-by-step explanation:

f'(x)=x^2-1/x^2

integrating both the sides

$f(x) = \frac{ {x}^{3} }{3} + \frac{1}{x} + c$

if f(1)=1/3

$\frac{ {(1)}^{3} }{3 } + \frac{1}{( 1)} + c = \frac{1}{3}$

c=-1

f(x)=(x^3)/3+(1/x)-1