<s>[INST] Find $ \int \frac{dx}{x\sqrt{1-x^4}}$ Find $\displaystyle \int \dfrac{dx}{x\sqrt{1-x^4}}$
I cannot figure out how start this problem, can anyone explain  
 [/INST] 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*}
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