[tex]f'(x)=2x-\dfrac3{x^4}[/tex]
[tex]\displaystyle\int f'(x)\,\mathrm dx=\int\left(2x-\frac3{x^4}\right)\,\mathrm dx[/tex]
[tex]f(x)=x^2+\dfrac1{x^3}+C[/tex]
Given that [tex]f(1)=2[/tex], we get
[tex]2=1^2+\dfrac1{1^3}+C[/tex]
[tex]2=2+C[/tex]
[tex]\implies C=2[/tex]
so that
[tex]f(x)=x^2+\dfrac1{x^3}[/tex]
