Question

Solve each of the integrals in Exercises 39–74. Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.

$\int \frac{x^2+9}{x^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}dx$

Short answer

The solution of the integral is$\frac{2}{3}{x}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{18}{\sqrt{x}}+C.$

Step-by-step solution

Step 1. Given Information.

The given integral is$\int \frac{x^2+9}{x^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}dx.$

Step 2. Solve. 

Let's solve the integral,

$$\begin{gathered} \int \frac{x^2+9}{x^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}dx \\ =\int \sqrt{x}+\frac{9}{x^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}dx \\ =\int \sqrt{x}dx+\int \frac{9}{x^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}dx \\ =\frac{2}{3}{x}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+9\left(-2\frac{1}{\sqrt{x}}\right)+C \\ =\frac{2}{3}{x}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{18}{\sqrt{x}}+C \end{gathered}$$