Integral $\int \frac{x}{\left(1-x^3\right)\sqrt{1-x^2}}dx$

Question

I thought this integral was simple, but it turns out it's not.

$$I=\int \frac{x}{\left(1-x^3\right)\sqrt{1-x^2}}\ dx$$

I tried the substitution $1-x^3=\frac{1}{t}$, but that leaves me again with $x = \sqrt{1-t^2}$, which doesn't simplify things. Then I tried $x = \sin(t)$, which gets me to: $$\int \frac{\sin t}{\left(1-\sin t\right)\left(1+\sin t+\sin^2 t\right)}dt$$

This looked solvable so I did another substitution: $\tan(\frac{t}{2})=u$, after a few steps I got this:

$$\int \:\frac{\left(u+u^3\right)du}{\left(u^2-2u+1\right)\left(2u^3+3u^2+4u+1\right)}$$

which is still too complicated. What am I missing? Any ideas?

Update: After splitting this function into

$$I=\frac{1}{3}\int \frac{1}{\left(1-x\right)\sqrt{1-x^2}}dx + \frac{1}{3}\int \frac{x-1}{\left(1+x+x^2\right)\sqrt{1-x^2}}dx$$

It gets a bit easier. I think I can solve the first integral. But, how do I solve the second?

Thank you!

Answer 1

After splitting the integral as follows \begin{align} &\int \frac{x}{\left(1-x^3\right)\sqrt{1-x^2}}\ dx \\ =& \ \frac{1}{3}\int \frac{1}{\left(1-x\right)\sqrt{1-x^2}}dx + \frac{1}{3}\int \frac{x-1}{\left(1+x+x^2\right)\sqrt{1-x^2}}dx \end{align} apply the Euler substitution ${\sqrt{1-x^2}}=({1+x})t$ to evaluate the two integrals

\begin{align} &\int \frac{1}{\left(1-x\right)\sqrt{1-x^2}}dx =-\int \frac1{t^2}dt=\frac1t\\ \\ &\int \frac{x-1}{\left(1+x+x^2\right)\sqrt{1-x^2}}dx\\ =&\int\frac{4t^2}{t^4+3}dt = 2\int\frac{t^2+\sqrt3}{t^4+3}+ \frac{t^2-\sqrt3}{t^4+3} \ dt\\ = &\frac{2}{\sqrt{2\sqrt3}}\tan^{-1}\frac{t^2-\sqrt3}{t\sqrt{2\sqrt3}} - \frac{2}{\sqrt{2\sqrt3}}\coth^{-1}\frac{t^2+\sqrt3}{t\sqrt{2\sqrt3}} \end{align}

Answer 2

Factorize the denominator as

$$(1-x^3)\sqrt{1-x^2}=(1-x)^\frac32(1+x)^\frac12(x^2+x+1)$$

So, the integral becomes

$$\mathcal I=\int\frac{\mathrm dx}{(1-x)^\frac32(1+x)^\frac12\left(x+\frac1{x}+1\right)}$$

Since the general method to integrate functions of the type $(x-\alpha)^m(x-\beta)^n$ where $\alpha, \beta, m, n\in\mathbb R$ such that $m+n=-2$ is to take one of $(x-\alpha)^2$ and $(x-\beta)^2$ common from the denominator and substitute for $\frac{x-\beta}{x-\alpha}$ or $\frac{x-\alpha}{x-\beta}$ respectively, this same approach can be extrapolated here.

So,

$$\mathcal I=\int\frac{\mathrm dx}{(1-x)^2\sqrt{\frac{1+x}{1-x}}\left(x+\frac1{x}+1\right)}$$

Setting $\require{cancel}t=\sqrt{\frac{1+x}{1-x}}$,

$$\cancel2t\mathrm dt=-\frac{\cancel2\mathrm dx}{(1-x)^2}, x=\frac{t^2+1}{t^2-1}$$ $$\begin{align}\implies\mathcal I&=\int\frac{1-t^4}{3t^4+1}\mathrm dt\\&=-\frac{t}3+\frac43\underbrace{\int\frac{\mathrm dt}{3t^4+1}}_{\mathcal J}\end{align}$$

For evaluating $\mathcal J$, substitute $u=\sqrt[4]3 t$ and then proceed by rewriting $2$ as $(1+u^2)-(1-u^2)$.

Answer 3

For the second integral, substitute $v_\pm=\left(\dfrac{\sqrt3\mp1}2\right)\dfrac{\sqrt{1-x^2}}{x+2\pm\sqrt3}:$

$$\begin{align}\int\frac{x-1}{(x^2+x+1)\sqrt{1-x^2}}\mathrm dx&=\frac{3\sqrt3-5}2\int\frac{\mathrm dv_+}{v_+^2-\frac{\left(2-\sqrt3\right)^2}{2\sqrt3}}+\frac{3\sqrt3+5}2\int\frac{\mathrm dv_-}{v_-^2+\frac{\left(2+\sqrt3\right)^2}{2\sqrt3}}\\&=\sqrt{\frac2{\sqrt3}}\left(\tan^{-1}\left(\frac{\sqrt{\frac{\sqrt3}2}\left(\sqrt3-1\right)\sqrt{1-x^2}}{x+2-\sqrt3}\right)-\tanh^{-1}\left(\frac{\sqrt{\frac{\sqrt3}2}\left(\sqrt3+1\right)\sqrt{1-x^2}}{x+2+\sqrt3}\right)\right)+C\end{align}$$