Question

$\int \frac{2xdx}{\sqrt{1-x^2-x^4}}dx$

Answer 1

We have,

$\int \frac{2xdx}{\sqrt{1-x^2-x^4}}$

$\Rightarrow \int \frac{2xdx}{\sqrt{1-x^2-{\left(x^2\right)}^2}}$

Suppose that,

$x^2=t$

$2xdx=dt$

$\int \frac{dt}{\sqrt{1-t-t^2}}$

Making completing square denominator.

$\int \frac{dt}{\sqrt{1-t-t^2+\frac{1}{4}-\frac{1}{4}}}$

$\Rightarrow \int \frac{dt}{\sqrt{1+\frac{1}{4}-t-t^2-\frac{1}{4}}}$

$\Rightarrow \int \frac{dt}{\sqrt{\frac{5}{4}-(t+t^2+\frac{1}{4})}}$

$\Rightarrow \int \frac{dt}{\sqrt{\frac{5}{4}-{(t+\frac{1}{2})}^2}}$

$\Rightarrow \int \frac{dt}{\sqrt{{\left(\frac{\sqrt{5}}{2}\right)}^2-{(t+\frac{1}{2})}^2}}$

Applying formula

$\int \frac{1}{\sqrt{a^2-x^2}}dx={\sin }^{-1}\frac{x}{a}$

Then,

$\Rightarrow \int \frac{dt}{\sqrt{{\left(\frac{\sqrt{5}}{2}\right)}^2-{(t+\frac{1}{2})}^2}}={\sin }^{-1}\frac{(t+\frac{1}{2})}{\left(\frac{\sqrt{5}}{2}\right)}$

Put $t=x^2$

$\int \frac{2xdx}{\sqrt{1-x^2-x^4}}={\sin }^{-1}\frac{x^2+\frac{1}{2}}{\frac{\sqrt{5}}{2}}+C$

$={\sin }^{-1}\frac{2x^2+1}{\sqrt{5}}+C$

Hence, this is the answer.