Answer
$$C\left( x \right) = \frac{{5{x^2}}}{2} - \ln \left| x \right| + \ln 10 - 155.8$$
Work Step by Step
$$\eqalign{
& C'\left( x \right) = 5x - 1/x;{\text{ 10 units cost }}\$ 94.20 \cr
& {\text{The marginal function cost is the derivative of the function cost }}C'\left( x \right) \cr
& {\text{then}}{\text{, the function cost }}C\left( x \right){\text{ is }} \cr
& C\left( x \right) = \int {C'\left( x \right)} dx \cr
& {\text{replacing }}5x - 1/x{\text{ for }}C'\left( x \right) \cr
& C\left( x \right) = \int {\left( {5x - \frac{1}{x}} \right)} dx \cr
& {\text{integrating }} \cr
& C\left( x \right) = \frac{{5{x^{1 + 1}}}}{{1 + 1}} - \ln \left| x \right| + K \cr
& C\left( x \right) = \frac{{5{x^2}}}{2} - \ln \left| x \right| + K{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& {\text{Find }}K,{\text{we know that the 10 units cost }}\$ 94.20{\text{ then }}C\left( {10} \right) = 94.20 \cr
& 94.20 = \frac{{5{{\left( {10} \right)}^2}}}{2} - \ln \left| {10} \right| + K \cr
& 94.20 = 250 - \ln 10 + K \cr
& K = \ln 10 + 94.20 - 250 \cr
& K = \ln 10 - 155.8 \cr
& {\text{then substituting }}K = \ln 10 - 155.8{\text{ into the equation }}\left( {\bf{1}} \right){\text{ to obtain}} \cr
& C\left( x \right) = \frac{{5{x^2}}}{2} - \ln \left| x \right| + \ln 10 - 155.8 \cr} $$
