Evaluate iterated integral. ∫12∫493+5yxdxdy

We can evaluate the double integral by integrating with respect to each variable one by one according to their order in the integration term . After integrating with respect to one variable , we have to apply its limit and then we can perform integration using next variable .
The following integration formulas can be used here ,
${\displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}+C}$
${\displaystyle \int kxdx=k\frac{x^2}{2}+C}$
${\displaystyle \int \frac{1}{\sqrt{x}}dx=\int x^{-\frac{1}{2}}dx}$
Consider the double integral ${\displaystyle {\int }_1^2{\int }_4^9\frac{3+5y}{\sqrt{x}}dxdy}$
Where ${\displaystyle \sqrt{x}}$ can be written as ${\displaystyle x^{\frac{1}{2}}}$ using the conversion formula of radical to exponent ${\displaystyle \sqrt{m}\left\{a^n\right\}=a^{\frac{n}{m}}}$
The expression can be written as ${\displaystyle (3+5y)x^{-\frac{1}{2}}}$ by taking ${\displaystyle x^{\frac{1}{2}}}$ to the numerator.
Thus we can write the integral as ${\displaystyle {\int }_1^2{\int }_4^9(3+5y)x^{-\frac{1}{2}}dxdy}$
First we have to integrate with respect to x by taking y terms outside as a constant.
Use the formula ${\displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}+C}$
${\displaystyle {\int }_1^2(3+5y)\left[{\int }_4^9{x}^{-\frac{1}{2}}dx\right]dy={\int }_1^2(3+5y)\left[{\left(\frac{x^{-\frac{1}{2}+1}}{\frac{-1}{2}+1}\right)}_4^9\right]dy}$
${\displaystyle ={\int }_1^2(3+5y)\left[{\left(\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right)}_4^9\right]dy}$
${\displaystyle ={\int }_1^2(3+5y)\left[2(9^{\frac{1}{2}}-4^{\frac{1}{2}})\right]dy}$