Integral of $$$\frac{x}{\sqrt{- 6 x^{4} + 2 x^{2}}}$$$
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Your Input
Find $$$\int \frac{x}{\sqrt{- 6 x^{4} + 2 x^{2}}}\, dx$$$.
Solution
The input is rewritten: $$$\int{\frac{x}{\sqrt{- 6 x^{4} + 2 x^{2}}} d x}=\int{\frac{1}{\sqrt{2 - 6 x^{2}}} d x}$$$.
Simplify the integrand:
$${\color{red}{\int{\frac{1}{\sqrt{2 - 6 x^{2}}} d x}}} = {\color{red}{\int{\frac{\sqrt{2}}{2 \sqrt{1 - 3 x^{2}}} d x}}}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{\sqrt{2}}{2}$$$ and $$$f{\left(x \right)} = \frac{1}{\sqrt{1 - 3 x^{2}}}$$$:
$${\color{red}{\int{\frac{\sqrt{2}}{2 \sqrt{1 - 3 x^{2}}} d x}}} = {\color{red}{\left(\frac{\sqrt{2} \int{\frac{1}{\sqrt{1 - 3 x^{2}}} d x}}{2}\right)}}$$
Let $$$x=\frac{\sqrt{3} \sin{\left(u \right)}}{3}$$$.
Then $$$dx=\left(\frac{\sqrt{3} \sin{\left(u \right)}}{3}\right)^{\prime }du = \frac{\sqrt{3} \cos{\left(u \right)}}{3} du$$$ (steps can be seen »).
Also, it follows that $$$u=\operatorname{asin}{\left(\sqrt{3} x \right)}$$$.
Therefore,
$$$\frac{1}{\sqrt{1 - 3 x^{2}}} = \frac{1}{\sqrt{1 - \sin^{2}{\left( u \right)}}}$$$
Use the identity $$$1 - \sin^{2}{\left( u \right)} = \cos^{2}{\left( u \right)}$$$:
$$$\frac{1}{\sqrt{1 - \sin^{2}{\left( u \right)}}}=\frac{1}{\sqrt{\cos^{2}{\left( u \right)}}}$$$
Assuming that $$$\cos{\left( u \right)} \ge 0$$$, we obtain the following:
$$$\frac{1}{\sqrt{\cos^{2}{\left( u \right)}}} = \frac{1}{\cos{\left( u \right)}}$$$
Integral becomes
$$\frac{\sqrt{2} {\color{red}{\int{\frac{1}{\sqrt{1 - 3 x^{2}}} d x}}}}{2} = \frac{\sqrt{2} {\color{red}{\int{\frac{\sqrt{3}}{3} d u}}}}{2}$$
Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=\frac{\sqrt{3}}{3}$$$:
$$\frac{\sqrt{2} {\color{red}{\int{\frac{\sqrt{3}}{3} d u}}}}{2} = \frac{\sqrt{2} {\color{red}{\left(\frac{\sqrt{3} u}{3}\right)}}}{2}$$
Recall that $$$u=\operatorname{asin}{\left(\sqrt{3} x \right)}$$$:
$$\frac{\sqrt{6} {\color{red}{u}}}{6} = \frac{\sqrt{6} {\color{red}{\operatorname{asin}{\left(\sqrt{3} x \right)}}}}{6}$$
Therefore,
$$\int{\frac{1}{\sqrt{2 - 6 x^{2}}} d x} = \frac{\sqrt{6} \operatorname{asin}{\left(\sqrt{3} x \right)}}{6}$$
Add the constant of integration:
$$\int{\frac{1}{\sqrt{2 - 6 x^{2}}} d x} = \frac{\sqrt{6} \operatorname{asin}{\left(\sqrt{3} x \right)}}{6}+C$$
Answer
$$$\int \frac{x}{\sqrt{- 6 x^{4} + 2 x^{2}}}\, dx = \frac{\sqrt{6} \operatorname{asin}{\left(\sqrt{3} x \right)}}{6} + C$$$A