Integral of $$$\sqrt{x y}$$$ with respect to $$$x$$$
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Your Input
Find $$$\int \sqrt{x y}\, dx$$$.
Solution
The input is rewritten: $$$\int{\sqrt{x y} d x}=\int{\sqrt{x} \sqrt{y} d x}$$$.
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\sqrt{y}$$$ and $$$f{\left(x \right)} = \sqrt{x}$$$:
$${\color{red}{\int{\sqrt{x} \sqrt{y} d x}}} = {\color{red}{\sqrt{y} \int{\sqrt{x} d x}}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=\frac{1}{2}$$$:
$$\sqrt{y} {\color{red}{\int{\sqrt{x} d x}}}=\sqrt{y} {\color{red}{\int{x^{\frac{1}{2}} d x}}}=\sqrt{y} {\color{red}{\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}=\sqrt{y} {\color{red}{\left(\frac{2 x^{\frac{3}{2}}}{3}\right)}}$$
Therefore,
$$\int{\sqrt{x} \sqrt{y} d x} = \frac{2 x^{\frac{3}{2}} \sqrt{y}}{3}$$
Add the constant of integration:
$$\int{\sqrt{x} \sqrt{y} d x} = \frac{2 x^{\frac{3}{2}} \sqrt{y}}{3}+C$$
Answer
$$$\int \sqrt{x y}\, dx = \frac{2 x^{\frac{3}{2}} \sqrt{y}}{3} + C$$$A