Integral of $$$\frac{x}{\sqrt{1 - 16 x^{2}}}$$$

Find $$$\int \frac{x}{\sqrt{1 - 16 x^{2}}}\, dx$$$.

Let $$$u=1 - 16 x^{2}$$$.

Then $$$du=\left(1 - 16 x^{2}\right)^{\prime }dx = - 32 x dx$$$ (steps can be seen »), and we have that $$$x dx = - \frac{du}{32}$$$.

So,

$${\color{red}{\int{\frac{x}{\sqrt{1 - 16 x^{2}}} d x}}} = {\color{red}{\int{\left(- \frac{1}{32 \sqrt{u}}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{1}{32}$$$ and $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$:

$${\color{red}{\int{\left(- \frac{1}{32 \sqrt{u}}\right)d u}}} = {\color{red}{\left(- \frac{\int{\frac{1}{\sqrt{u}} d u}}{32}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=- \frac{1}{2}$$$:

$$- \frac{{\color{red}{\int{\frac{1}{\sqrt{u}} d u}}}}{32}=- \frac{{\color{red}{\int{u^{- \frac{1}{2}} d u}}}}{32}=- \frac{{\color{red}{\frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}}{32}=- \frac{{\color{red}{\left(2 u^{\frac{1}{2}}\right)}}}{32}=- \frac{{\color{red}{\left(2 \sqrt{u}\right)}}}{32}$$

Recall that $$$u=1 - 16 x^{2}$$$:

$$- \frac{\sqrt{{\color{red}{u}}}}{16} = - \frac{\sqrt{{\color{red}{\left(1 - 16 x^{2}\right)}}}}{16}$$

Therefore,

$$\int{\frac{x}{\sqrt{1 - 16 x^{2}}} d x} = - \frac{\sqrt{1 - 16 x^{2}}}{16}$$

Add the constant of integration:

$$\int{\frac{x}{\sqrt{1 - 16 x^{2}}} d x} = - \frac{\sqrt{1 - 16 x^{2}}}{16}+C$$

$$$\int \frac{x}{\sqrt{1 - 16 x^{2}}}\, dx = - \frac{\sqrt{1 - 16 x^{2}}}{16} + C$$$A