Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 275: 33

$$ \sqrt{x^{2}+9 }+C$$

Given $$\int \frac{x}{\sqrt{x^{2}+9}} d x $$ Let $$ u= x^{2}+9\ \ \ \Rightarrow \ \ \ du=(2x)dx$$ Then \begin{aligned} \int \frac{x}{\sqrt{x^{2}+9}} d x &=\frac{1}{2} \int \frac{d u}{\sqrt{u}} \\ &=\frac{1}{2} \int u^{-1 / 2} d u \\ &=\frac{1}{2}\left(2 u^{1 / 2}+C\right) \\ &=u^{1 / 2}+C \\ &= \sqrt{x^{2}+9 }+C\end{aligned}