Answer:
C(x)=[tex]9x+\frac{1}{2(x^2+1)}+1081.9[/tex]
Step-by-step explanation:
We are given that
[tex]C'(x)=9-\frac{x}{(x^2+1)^2}[/tex]
Total cost function
C(x)=[tex]\int(9-\frac{x}{(x^2+1)^2})dx[/tex]
Let [tex]x^2+1=t[/tex]
[tex]2xdx=dt[/tex]
[tex]xdx=\frac{1}{2}dt[/tex]
[tex]\int\frac{x}{(x^2+1)^2}dx=\int\frac{1}{2t^2}dt=-\frac{1}{2t}=-\frac{1}{2(x^2+1)}[/tex]
Substitute the value
[tex]C(x)=9x+\frac{1}{2(x^2+1)}+C[/tex]
Substitute x=2
[tex]C(2)=9(2)+\frac{1}{2(5)}+C[/tex]
[tex]1100=18+\frac{1}{10}+C[/tex]
[tex]1100-18-\frac{1}{10}=C[/tex]
[tex]1082-0.1=C[/tex]
[tex]1081.9=C[/tex]
Total cost function, C(x)=[tex]9x+\frac{1}{2(x^2+1)}+1081.9[/tex]