Integral of $$$\frac{5 x}{\left(4 - 3 x^{2}\right)^{3}}$$$

Find $$$\int \frac{5 x}{\left(4 - 3 x^{2}\right)^{3}}\, dx$$$.

Let $$$u=4 - 3 x^{2}$$$.

Then $$$du=\left(4 - 3 x^{2}\right)^{\prime }dx = - 6 x dx$$$ (steps can be seen »), and we have that $$$x dx = - \frac{du}{6}$$$.

Thus,

$${\color{red}{\int{\frac{5 x}{\left(4 - 3 x^{2}\right)^{3}} d x}}} = {\color{red}{\int{\left(- \frac{5}{6 u^{3}}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{5}{6}$$$ and $$$f{\left(u \right)} = \frac{1}{u^{3}}$$$:

$${\color{red}{\int{\left(- \frac{5}{6 u^{3}}\right)d u}}} = {\color{red}{\left(- \frac{5 \int{\frac{1}{u^{3}} d u}}{6}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-3$$$:

$$- \frac{5 {\color{red}{\int{\frac{1}{u^{3}} d u}}}}{6}=- \frac{5 {\color{red}{\int{u^{-3} d u}}}}{6}=- \frac{5 {\color{red}{\frac{u^{-3 + 1}}{-3 + 1}}}}{6}=- \frac{5 {\color{red}{\left(- \frac{u^{-2}}{2}\right)}}}{6}=- \frac{5 {\color{red}{\left(- \frac{1}{2 u^{2}}\right)}}}{6}$$

Recall that $$$u=4 - 3 x^{2}$$$:

$$\frac{5 {\color{red}{u}}^{-2}}{12} = \frac{5 {\color{red}{\left(4 - 3 x^{2}\right)}}^{-2}}{12}$$

Therefore,

$$\int{\frac{5 x}{\left(4 - 3 x^{2}\right)^{3}} d x} = \frac{5}{12 \left(4 - 3 x^{2}\right)^{2}}$$

Simplify:

$$\int{\frac{5 x}{\left(4 - 3 x^{2}\right)^{3}} d x} = \frac{5}{12 \left(3 x^{2} - 4\right)^{2}}$$

Add the constant of integration:

$$\int{\frac{5 x}{\left(4 - 3 x^{2}\right)^{3}} d x} = \frac{5}{12 \left(3 x^{2} - 4\right)^{2}}+C$$

$$$\int \frac{5 x}{\left(4 - 3 x^{2}\right)^{3}}\, dx = \frac{5}{12 \left(3 x^{2} - 4\right)^{2}} + C$$$A