integration for |x|<1 for a simple integral I want to integrate 

$$\int_{-1}^{1}\frac{dx}{x^3\sqrt{1-x^2}}.$$

I'm not sure where to begin as I have tried integrating by parts but end up in a continuous circle
 A: Hint: Try the substitution $u=\sqrt{1-x^2}$.
A: The way I would approach this integral is by using trigonometric substitution
$$\int_{a}^{b}\frac{dx}{x^3\sqrt{1-x^2}}.$$
$$x=sin(θ)$$
$$dx=cos(θ)dθ$$
$$\int_{a}^{b}\frac{dθ}{sin^3(θ)\sqrt{1-sin^2(θ)}}.$$
you can simplify $$1-sin^2(θ)=cos^2(θ)$$
$$√(cos^2(θ))=cos(θ)$$
$$\int_{a}^{b}\frac{dθ}{sin^3(θ)\cos(θ)}.$$
Well this is the same as 
$$\int_{a}^{b}csc^3(θ)dθ.$$
The is a simple integration by parts.
$$\int_{a}^{b}udv=uv|_{a}^{b}-\int_{a}^{b}vdu.$$
so we have to step up the substitutions again.
$$u=csc(θ)$$
$$du=-csc(θ)cot(θ)$$
$$dv=csc^2(θ)$$
$$v=-cot(θ)$$
$$\int_{a}^{b}csc^3(θ)dθ=-csc(θ)cot(θ)|_{a}^{b}-\int_{a}^{b}(-cot(θ))(-csc(θ)cot(θ))dθ.$$
$$\int_{a}^{b}csc^3(θ)dθ=-csc(θ)cot(θ)|_{a}^{b}-\int_{a}^{b}(csc(θ)cot^2(θ))dθ.$$
we could rewrite $$cot^2(θ)=csc^2(θ)-1$$ because we know that $$1+cot^2(θ)=csc^2(θ)$$
$$\int_{a}^{b}csc^3(θ)dθ=-csc(θ)cot(θ)|_{a}^{b}-\int_{a}^{b}(csc(θ)(csc^2(θ)-1))dθ.$$
$$\int_{a}^{b}csc^3(θ)dθ=-csc(θ)cot(θ)|_{a}^{b}-\int_{a}^{b}((csc^3(θ)-csc(θ))dθ.$$
we can just add the intergral of $$csc^3(θ)$$ to the left side and we get 2.
$$\int_{a}^{b}csc^3(θ)dθ=-csc(θ)cot(θ)|_{a}^{b}-\int_{a}^{b}(csc^3(θ))+\int_{a}^{b}csc(θ)dθ.$$
$$2\int_{a}^{b}csc^3(θ)dθ=-csc(θ)cot(θ)|_{a}^{b}+\int_{a}^{b}csc(θ)dθ.$$
so we can just the 2 from the left side. and Voilà. we get the $$\int_{a}^{b}csc^3(θ)dθ$$ 
$$\int_{a}^{b}csc^3(θ)dθ=\frac{1}{2}(-csc(θ)cot(θ))|_{a}^{b}+\frac{1}{2}\int_{a}^{b}csc(θ)dθ.$$.
$$\frac{1}{2}\int_{a}^{b}csc(θ)dθ-\frac{1}{2}csc(θ)cot(θ))|_{a}^{b}$$.
we make my integration a bit simpler i am going to make an other substitution.
but before that i am going to multiply numerator and denominators of $$csc(θ)$$ by $$cot(θ)+csc(θ)$$. Which will give us.
$$\frac{1}{2}\int_{a}^{b}-\frac{-csc^2(θ)-cot(θ)csc(θ)}{cot(θ)+csc(θ)}dθ-\frac{1}{2}csc(θ)cot(θ))|_{a}^{b}$$.
Now i am going to substitute $$ z=cot(θ)+csc(θ) $$ and $$ dz=-csc^2(θ)-cot(θ)csc(θ)dθ $$
$$-\frac{1}{2}\int_{a}^{b}\frac{1}{z}dz-\frac{1}{2}csc(θ)cot(θ))|_{a}^{b}$$.
this integral is going to result to ->
$$-\frac{1}{2}\ln(z)|_{a}^{b}-\frac{1}{2}csc(θ)cot(θ))|_{a}^{b}$$.
then we can back substitute  z=$$cot(θ)+csc(θ)$$
which will give us. 
$$-\frac{1}{2}\ln(cot(θ)+csc(θ))|_{a}^{b}-\frac{1}{2}csc(θ)cot(θ))|_{a}^{b}$$.
Then we can back substitute θ as well.
$$θ=arcsin(x)$$
Then we get.
$$-\frac{\sqrt{1-x^2}}{2x^2}|_{a}^{b}-\frac{1}{2}\ln\frac{\sqrt{1-x^2}+1}{x}|_{a}^{b} $$
we can simplify this to
$$ -\frac{\sqrt{1-x^2}+x^2ln\frac{\sqrt{1-x^2}+1}{x}}{2x^2}|_{a}^{b} $$
and we can simplify this further more.
$$ -\frac{\sqrt{1-x^2}+x^2(arcsech(x))}{2x^2}|_{a}^{b} $$
the final answer is.
$$ \frac{1}{2}(\frac{-\sqrt{1-x^2})}{x^2}-ln(\sqrt{1-x^2}+1)+ln(x))|_{a}^{b} $$
just plug in b and a . minus it from each other. and thats the final answer
$$ \frac{1}{2}((\frac{-\sqrt{1-b^2})}{b^2}-ln(\sqrt{1-b^2}+1)+ln(b))-(\frac{-\sqrt{1-a^2})}{a^2}-ln(\sqrt{1-a^2}+1)+ln(a))) $$
A: Use $x=\sqrt{1-u^2}$
$$
\begin{align}
\int\frac{\mathrm{d}x}{x^3\sqrt{1-x^2}}
&=\int\frac{\mathrm{d}\sqrt{1-u^2}}{\sqrt{1-u^2}^3u}\\
&=-\int\frac{\mathrm{d}u}{(1-u^2)^2}\\
&=-\frac14\int\left(\frac1{(1-u)^2}+\frac1{(1+u)^2}+\frac1{1+u}+\frac1{1-u}\right)\,\mathrm{d}u\\
&=C-\frac14\left(\frac1{1-u}-\frac1{1+u}+\log(1+u)-\log(1-u)\right)\\
&=C-\frac14\left(\frac{2u}{1-u^2}+2\log\left(\frac{1+u}{\sqrt{1-u^2}}\right)\right)\\
&=C-\frac12\left(\frac{\sqrt{1-x^2}}{x^2}+\log\left(\frac{1+\sqrt{1-x^2}}{x}\right)\right)\tag{1}
\end{align}
$$
The integral above was done assuming that $x\gt0$. However, this can easily be extended by noticing that the integrand is an odd function, therefore, the integral is an even function. That is, the following is valid for all $x$:
$$
\int\frac{\mathrm{d}x}{x^3\sqrt{1-x^2}}
=C-\frac12\left(\frac{\sqrt{1-x^2}}{x^2}+\log\left(\frac{1+\sqrt{1-x^2}}{|x|}\right)\right)\tag{2}
$$
If your domain of integration spans the origin, the integral won't converge absolutely, but the Cauchy Principal Value will be given by $(2)$.

The question has been changed so that the domain of integration spans the origin. As mentioned above, the integral now does not converge absolutely. However, the Cauchy Principal Value is $0$.
