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

Question

Find $\displaystyle \int \dfrac{dx}{x\sqrt{1-x^4}}$

I cannot figure out how start this problem, can anyone explain

Answer 1

Start with $u=x^2$. Then $du=2xdx$, $dx=du/2x$, so we have

$$\int \frac{du}{2u \sqrt{1-u^2}}$$

From here you can perform a trigonometric substitution. Consider a right triangle whose hypotenuse is length $1$, such that one side has length $u$. Then the other side has length $\sqrt{1-u^2}$. Suppose $\theta$ is opposite the side of length $u$. Then $\sin(\theta)=u$ and $\cos(\theta)=\sqrt{1-u^2}$. Thus $du=\cos(\theta) d \theta$, so we have

$$\int \frac{\cos(\theta) d \theta}{2 \sin(\theta) \cos(\theta)} = \frac{1}{2} \int \csc(\theta) d \theta$$

This last integral is fairly standard, so I will leave it and the back-substitution to you.

Answer 2

Hints; Substitute $$u=x^4$$ then $$v=1-u$$ then $$m=\sqrt{v}$$

Answer 3

The substitution $u=\sqrt{1-x^{4}}$ converts the integral into a rational function of $u$. Since $x^{4}=1-u^{2}$ and $du=-\frac{2x^{3}}{u}dx$, we get

\begin{equation*} I=\int \frac{dx}{x\sqrt{1-x^{4}}}=\int \frac{1}{xu}\left( -\frac{u}{2x^{3}}% \right) du=\frac{1}{2}\int \frac{1}{u^{2}-1}\,du. \end{equation*}

Expanding the integrand into partial fractions \begin{equation*} \frac{1}{u^{2}-1}=\frac{1}{2\left( u-1\right) }-\frac{1}{2\left( u+1\right) } \end{equation*}

the integral is expressable in terms of logarithms \begin{equation*} I=\frac{1}{4}\int \frac{du}{u-1}-\frac{1}{4}\int \frac{du}{u+1}=\frac{1}{4} \ln \left( \left\vert \frac{\sqrt{1-x^{4}}-1}{\sqrt{1-x^{4}}+1}\right\vert \right) +C. \end{equation*}

Alternatively, using the standard integral \begin{equation*} \int \frac{1}{1-u^{2}}\,du=\operatorname{artanh}u+C \end{equation*}

we obtain an inverse hyperbolic function \begin{equation*} I=-\frac{1}{2}\operatorname{artanh}\sqrt{1-x^{4}}+C. \end{equation*}

Answer 4

$$ \begin{aligned} \int\frac{\mathrm{d}x}{x\sqrt{1-x^4}}&=\int\frac{\mathrm{d}x}{x^3\sqrt{1/x^4 - 1}}\\ &=\int\frac{1}{\sqrt{(1/x^2)^2 - 1}}\frac{\mathrm{d}x}{x^3}\\ &=\int\frac{1/x^2+\sqrt{(1/x^2)^2 - 1}}{\sqrt{(1/x^2)^2 - 1}}\,\frac{1}{1/x^2 + \sqrt{(1/x^2)^2 - 1}}\frac{\mathrm{d}x}{x^3}\\ &=-\frac{1}{2}\int \frac{1}{1/x^2 + \sqrt{(1/x^2)^2 - 1}}\left(1 + \frac{1/x^2}{\sqrt{(1/x^2)^2 - 1}}\right)\frac{-2}{x^3}\,\mathrm{d}x \end{aligned} $$

Set $t = 1/x^2 + \sqrt{(1/x^2)^2 - 1}$, so $\mathrm{d}t=\left(1 + \frac{1/x^2}{\sqrt{(1/x^2)^2 - 1}}\right)\frac{-2}{x^3}\,\mathrm{d}x$. Then $$ \int\frac{\mathrm{d}x}{x\sqrt{1-x^4}}=-\frac{1}{2}\log\left(\frac{1+\sqrt{1-x^4}}{x^2}\right) + C. $$