Integral of $$$\frac{x}{\sqrt{- 6 x^{4} + 2 x^{2}}}$$$

The calculator will find the integral/antiderivative of $$$\frac{x}{\sqrt{- 6 x^{4} + 2 x^{2}}}$$$, with steps shown.

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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


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